Trigonometry

1. Trigonometric ratios

Finding a trigonometric ratio

Answer the following questions about this right-angled triangle:

Complete the definition of $\cos \theta$ using the dropdown menus below:

Answer:
$\cos \theta$ is defined as the over the .

HINT: <no title>
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You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Complete the definition of $\cos \theta$
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We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[0.2cm] \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}$

For a right-angled triangle, the side across the triangle from the right-angle is called the "hypotenuse". For either of the other angles in the triangle, the side across the triangle from the angle is labelled as the "opposite". And the side next to this angle is called the "adjacent" side.

Therefore, $\cos \theta$ is defined as the adjacent over the hypotenuse.

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What is the value of the ratio $\cos \theta$ for the given right-angled triangle?

Answer: $\cos \theta =$ .
numeric

HINT: <no title>
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Identify which two sides of the triangle correspond to the sides you used in the definition of $\cos \theta$ in Question 1, and take the ratio of these two sides.

STEP: Take the ratio of the correct two sides of the triangle
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This question asks us to determine $\cos \theta$ for the given right-angled triangle.

From Question 1, we know that $\cos \theta$ is defined as the adjacent over the hypotenuse. We need to identify which sides of the triangle are the adjacent and hypotenuse, and take the ratio of their lengths.

For the given triangle, the adjacent has length $\text{3}$ and the hypotenuse has length $\text{5}$ .

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ = \frac{3}{5} \end{aligned}$ $\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \\[0.2cm] & = \frac{3}{5} \end{aligned}$

Therefore, $\cos \theta$ is $\dfrac{ 3 }{ 5 }$ .

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A connection between sine and cosine

The diagram below shows a right-angled triangle, with acute angles labelled $x$ and $y$ , $x \neq y$ . The sides of the triangle are labelled $A$ , $B$ , and $C$ .

Because this is a right-angled triangle, the following statement is true:

\begin{matrix} \sin x = \cos y \end{matrix}

$\begin{array}{c}\sin x = \cos y\end{array}$

Why is this statement true?

Answer:

$\sin x = \cos y$ because .

HINT: <no title>
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Identify the sides of the triangle needed for the ratios $\sin x$ and $\cos y$ . What do these sides have in common?

STEP: Identify the sides used for

\sin x

$\sin x$ and

\cos y

$\cos y$
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The trigonometric ratios sine ( $\sin$ ) and cosine ( $\cos$ ) tell us about the ratio of different sides relative to a particular angle $\theta$ in a right-angled triangle.

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \end{aligned}$

Deciding which side of the triangle is labelled as the 'opposite' side or the 'adjacent' side depends on the location of the angle that we are working with. The opposite side is always directly across the triangle from the angle. The adjacent side is the side next to the angle. (The hypotenuse is always the side across from the right-angle.)

For $\sin x$ , we need to identify the side opposite the angle $x$ . For $\cos y$ , we need to find the side adjacent to the angle $y$ . This is shown with the red and blue arrows below.

Looking at the picture, we can see that the arrows are pointing to the same side! Since this side is the numerator of both of the ratios, and the hypotenuse is the denominator for both of them, we now know

\begin{aligned} \sin x & = \frac{B}{C} \\ = \cos y \end{aligned}

$\begin{aligned} \sin x & = \frac{B}{C} \\ & = \cos y \end{aligned}$

This rule is always true for complementary angles (angles that add up to

90 °

$90°$ ) in a right-angled triangle. This is also why the cosine ratio is called cosine: it is the COmplementary function of SINE!

Therefore, the correct answer is $\sin x = \cos y$ because the side opposite $x$ is the side adjacent to $y$ .

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Opposite and adjacent sides: connecting to angles

The diagram below shows a right-angled triangle with sides $Q$ , $R$ , and $S$ , and two acute angles: $\omega$ , and $\theta$ .

Which side is adjacent to the angle $\theta$ ?

Answer: The side adjacent to angle

θ

$\theta$ is .

HINT: <no title>
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The side that is adjacent to the angle $\theta$ is situated next to the angle $\theta$ .

STEP: Identify the side adjacent to the angle

θ

$\theta$
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The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

The side that does not touch the $\text{90}\text{°}$ angle is the hypotenuse. The hypotenuse is also the longest side of a right-angled triangle. In this question, the hypotenuse is the side $S$ .

We label the other two sides according to the location of the angle of interest. This question asked about the sides adjacent and opposite to the angle $\theta$ , so the angle of interest is angle $\theta$ . Using this angle, we label the other two sides in this manner:

The side that touches the angle $\theta$ is the adjacent side.
The remaining side that is faraway from the angle $\theta$ is the opposite side.

There is another acute angle in the triangle, which we can call $\omega$ . If we change the angle of interest to $\omega$ , the labels opposite and adjacent will change. However, the hypotenuse remains the same.

We can summarise the above information using the diagram below:

NOTE: The sides

R

$R$ and

Q

$Q$ can each be the opposite side or the adjacent side. It depends on the angle of interest.

The side adjacent to angle $\theta$ is $R$ .

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Working with trigonometric ratios

Determine the value of $\tan \theta$ in the following triangle.

Select an answer from the following list:

A	$\dfrac{\sqrt{3}}{2}$
B	$\dfrac{\sqrt{2}}{2}$
C	$\dfrac{1}{2}$
D	$1$
E	$\dfrac{\sqrt{3}}{3}$
F	$\sqrt{3}$

Answer:

\tan θ =

$\tan \theta =$

HINT: <no title>
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Identify the opposite, adjacent, and hypotenuse sides of the triangle. Then use the definition of $\tan \theta$ to get the answer.

STEP: Use the definition of

\tan θ

$\tan \theta$ to get the answer
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We have been given the lengths of the three sides of a right-angled triangle. We need to determine the value of $\tan \theta$ . The definition of the tangent ratio is:

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \end{aligned}

$\begin{aligned} \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}$

We need to find the lengths for the opposite side and the adjacent side. The opposite and the adjacents sides both depend on the location of angle $\theta$ . The purple arrows in the diagram below show those two sides. The opposite side is across the triangle from $\theta$ , while the adjacent side is next to $\theta$ .

In this case the length of the opposite side is $4 \sqrt{3}$ and the length of the adjacent side is $4$ . Therefore,

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ = \frac{4 \sqrt{3}}{4} \\ = \sqrt{3} \end{aligned}

$\begin{aligned} \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \\ & = \dfrac{4 \sqrt{3}}{4} \\ & = \sqrt{3} \end{aligned}$

NOTE: The hypotenuse of the triangle is greyed out in the diagram above. That is because it is not needed for this question: the tangent ratio does not care about that side of the triangle.

The correct choice from the list is F.

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Right triangles: opposite and adjacent sides

The diagram below shows a right-angled triangle with sides labelled Side $\text{1}$ , Side $\text{2}$ , and Side $\text{3}$ . There is also an acute angle $\theta$ .

Which sides are adjacent and opposite to the angle $\theta$ ?

Answer:

The side adjacent to angle $\theta$ is .

The side opposite to angle $\theta$ is .

HINT: <no title>
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The side that is adjacent to the angle $\theta$ is next to the angle $\theta$ . The side that is opposite to the angle $\theta$ is across from the angle $\theta$ .

STEP: Identify the sides opposite and adjacent to the angle

θ

$\theta$
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The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

The side that does not touch the $\text{90}\text{°}$ angle is the hypotenuse. The hypotenuse is also the longest side of a right-angled triangle. In this question, the hypotenuse is Side $\text{2}$ .

The side that touches the angle $\theta$ is the adjacent side.
The remaining side that is far away from the angle $\theta$ is the opposite side.

There is another acute angle in the triangle, which we can call $\phi$ . If we change the angle of interest to $\phi$ , the labels opposite and adjacent will change. However, the hypotenuse remains the same.

We can summarise the above information using the diagram below:

NOTE: Side

3

$\text{3}$ and Side

1

$\text{1}$ can each be the opposite side or the adjacent side. It depends on the angle of interest.

The side adjacent to angle $\theta$ is Side $\text{3}$ .

The side opposite to angle $\theta$ is Side $\text{1}$ .

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Calculation of the area of a triangle

A triangle $ABC$ is shown in the figure below. The length of $\overline{AC}$ is $\text{4}$ units. The size of the angle $B\hat{A}C$ is $\text{50}°$ .

Calculate the lengths of the line segments $\overline{AB}$ and $\overline{BC}$ . You will need to use your knowledge of trigonometric ratios for these calculations. Round your answers to 2 decimal places.
Answer:
$\overline{AB} =$ units
$\overline{BC} =$ units

numeric
numeric
HINT: <no title>
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Use the definitions of the sine and cosine functions.

STEP: Choose the correct trigonometric ratios
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We are given the following information:
- The triangle has a right angle ( $A\hat{B}C$ ) (shown in the diagram)
- $\overline{AC} = \text{4}$ units
- $B\hat{A}C = \text{50}°$
This means we can use the cosine ( $\cos$ ) and sine ( $\sin$ ) functions to detemine the lengths of the line segments $\overline{AB}$ and $\overline{BC}$ . The $\cos$ and $\sin$ of an angle, for example, $\theta$ are defined as follows:

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \sin θ & = \frac{opposite}{hypotenuse} \end{aligned}$ $\begin{aligned} \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$

Line segment $\overline{AB}$ is adjacent to the angle $B\hat{A}C$ and line segment $\overline{BC}$ is opposite to the angle $B\hat{A}C$ . Line segment $\overline{AC}$ is the hypotenuse (it is opposite the right angle $A\hat{B}C$ ). We know the length of $\overline{AC}$ and the size of angle $B\hat{A}C$ , so we can calculate the unkown line segments.

Therefore, we will use the cosine function to calculate the length of $\overline{AB}$ and the sine function to calculate the length of $\overline{BC}$ .

STEP: Use the trigonometric ratios to find the sides of the triangle
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We will now calculate the lengths of $\overline{AB}$ and $\overline{BC}$ .

For $\overline{AB}$ :

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (B \hat{A} C) & = \frac{\bar{A B}}{\bar{A C}} \\ ∴ \bar{A B} & = \cos (B \hat{A} C) \times \bar{A C} \\ = \cos 50 ° \times 4 \\ = 2,57115... \\ \approx 2,57 units \end{aligned}$ $\begin{aligned} \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \cos (B\hat{A}C) & = \dfrac{\phantom{=} \overline{AB} \phantom{=}}{\overline{AC}} \\ \therefore \overline{AB} & = \cos (B\hat{A}C) \times \overline{AC} \\ & = \cos \text{50}° \times \text{4} \\ & = \text{2,57115...} \\ & \approx \text{2,57} \text{ units} \end{aligned}$

For $\overline{BC}$ :

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin (B \hat{A} C) & = \frac{\bar{B C}}{\bar{A C}} \\ ∴ \bar{B C} & = \sin (B \hat{A} C) \times \bar{A C} \\ = \sin 50 ° \times 4 \\ = 3,06417... \\ \approx 3,06 units \end{aligned}$ $\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \sin (B\hat{A}C) & = \dfrac{\phantom{=} \overline{BC} \phantom{=}}{\overline{AC}} \\ \therefore \overline{BC} & = \sin (B\hat{A}C) \times \overline{AC} \\ & = \sin \text{50}° \times \text{4} \\ & = \text{3,06417...} \\ & \approx \text{3,06} \text{ units} \end{aligned}$

Therefore, the length of $\overline{AB} = \text{2,57}$ units and the length of $\overline{BC} = \text{3,06}$ units.

Submit your answer as: and

Which formula from the table below would you use to calculate the area of the triangle $ABC$ ?

Choice	Formula
A	$\overline{AB} \times \overline{BC} \times \overline{CA}$
B	$\frac{1}{2} \times \overline{AB} \times \overline{BC} \times \overline{CA}$
C	$\overline{AB} \times \overline{BC}$
D	$\frac{1}{2} \times \overline{AB} \times \overline{BC}$
E	$\pi \times \left(\overline{AB}\right)^2$

Answer: The correct formula is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Consider a right triangle to be a result of dividing a rectangle into two parts along the diagonal. Do you remember the formula for the area of a rectangle?

STEP: Determine the formula using the relationship between a right triangle and a rectangle
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We can consider the triangle $ABC$ part of a rectangle $ABCD$ , as shown below:

The area of the rectangle $ABCD$ is simply length multiplied by the breadth, i.e. $\overline{AB} \times \overline{BC}$ or $\overline{CD} \times \overline{AD}$ . Notice, the area of the triangle $ABC$ is exactly half of the area of the rectangle $ABCD$ . Therefore the area of the triangle $ABC$ is equal to $\frac{1}{2} \times \overline{AB} \times \overline{BC}$ .

You can determine the area of any right angled triangle by multiplying the two sides adjacent to the right angle, and dividing by 2 (or multiplying by a half).

Therefore, the correct formula is choice D.

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Using the unrounded answers from the first question, calculate the area of the triangle $ABC$ . Round your answers to 2 decimal places.

Answer: Area of triangle $ABC =$ $\text{units}^2$
numeric

HINT: <no title>
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Use the formula for the area of a triangle.

STEP: Use the formula for the area of a triangle
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From the previous question, we have found that the formula for the area of the triangle $ABC$ is $\frac{1}{2} \times \overline{AB} \times \overline{BC}$ .

From Question 1, we found that $\overline{AB} = \text{2,57115...}$ units and $\overline{BC} = \text{3,06417...}$ units.

The area of the triangle $ABC$ :

$\begin{aligned} Area of triangle A B C & = \frac{1}{2} \times \bar{A B} \times \bar{B C} \\ = \frac{1}{2} \times 2,57115... \times 3,06417... \\ = 3,93923... \\ \approx 3,94 {units}^{2} \end{aligned}$ $\begin{aligned} \text{Area of triangle } ABC & = \frac{1}{2} \times \overline{AB} \times \overline{BC} \\ & = \frac{1}{2} \times \text{2,57115...} \times \text{3,06417...} \\ & = \text{3,93923...} \\ & \approx \text{3,94} \text{ units}^2 \end{aligned}$

Therefore, the area of the triangle $ABC$ is $\text{3,94}$ units $^2$ .

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Identify and use the correct trigonometric ratio

Study the given right-angled triangle $\Delta NSQ$ with $\angle N = \text{90} \text{°}$ and answer the following question. The angle given in the diagram is: $\angle NSQ = \text{24,66}{°}$ . The side given is: $\overline{ SQ } = \text{11}$ .

Calculate the length of side $\overline{SN}$ . Round your answer to two decimal places if appropriate.

Answer:

\bar{S N} =

$\overline{ SN } =$

numeric

HINT: <no title>
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The first thing you need to do is figure out which trigonometric ratio to use: sine, cosine, or tangent.

STEP: Find the correct trigonometric ratio to solve for the unkown
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From the question we know the side $\overline{ SQ } = \text{11}$ and the angle $\angle NSQ = \text{24,66}\text{°}$ . And we need to find $\overline{ SN }$ . These sides are the hypotenuse and the side adjacent to angle $\angle NSQ$ . The trigonometric ratio which links the adjacent side to the hypotenuse is cosine. We can write an equation based on the cosine ratio as show here, and we can call the unknown side $y$ .

\begin{aligned} \cos 24,66 ° & = \frac{\bar{S N}}{\bar{S Q}} \\ \cos 24,66 ° & = \frac{y}{\bar{S Q}} \end{aligned}

$\begin{aligned} \cos \text{24,66}{°} &= \frac{\overline{ SN }}{\overline{ SQ }}\\\cos \text{24,66}{°} &= \frac{y}{\overline{ SQ }} \end{aligned}$

STEP: Solve the trigonometric equation
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Now multiply both sides of the equation by the denominator.

\begin{aligned} \cos 24,66 ° & = \frac{y}{11,0} \\ 11,0 \cdot \cos 24,66 ° & = y \\ y & = 11,0 \cdot \cos 24,66 ° \\ y & = 9,99679... \\ y & \approx 10 \end{aligned}

$\begin{aligned} \cos \text{24,66}\text{°} &= \frac{y}{\text{11,0}}\\ \text{11,0} \cdot \cos \text{24,66}\text{°} &= y\\ y &= \text{11,0} \cdot \cos \text{24,66}\text{°} \\ y &= \text{9,99679...}\\ y &\approx \text{10} \end{aligned}$

Remember that the instructions say to round the answer to the second decimal place, as shown in the last step above.

The length of $\overline{ SN }$ is $\text{10}$ .

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Trigonometry: using a calculator

Use your calculator to evaluate the following expression: $\; \cos5°$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

\cos 5 ° \approx

$\cos5° \approx$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate $\cos5°$ . If you are not sure about how to do that, refer to the manual of your calculator.

STEP: Use your calculator to evaluate

\cos 5 °

$\cos5°$ and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

\begin{aligned} \cos 5 ° & = 0,99619... \\ \approx 1 \end{aligned}

$\begin{aligned} \cos5° & = \text{0,99619...} \\ & \approx \text{1} \end{aligned}$

The correct answer is $\text{1}$ .

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Trigonometric ratios

Which trigonometric ratio is defined as the ratio of the opposite and the hypotenuse?

Answer: The trigonometric ratio is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Identify the correct trigonometric ratio
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The trigonometric ratios tell us about two sides and an angle in a right-angled triangle.

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

$\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[0.2cm] \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}$

For a right-angled triangle, the side across the triangle from the right-angle is called the "hypotenuse". For either of the other angles in the triangle, the side across the triangle from the angle is labelled as the "opposite". And the side next to this angle is called the "adjacent" side.

The diagram shows the two sides we need for the sine ratio in purple: opposite and hypotenuse. The adjacent is shaded out because it is not needed for the sine ratio.

NOTE: These ratios are only valid when the triangle has a right-angle. Otherwise the triangle would not have a hypotenuse. Always check that you have a right-angled triangle before you start using these trigonometric ratios, otherwise you will not get the right answer.

Therefore, the ratio defined as the ratio of the opposite and the hypotenuse is sine.

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Trigonometric ratios: definitions

Complete the definition of sin using the dropdown menus below:

Answer:

Sin is defined as the over the .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Complete the definition of sin
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The trigonometric ratios tell us about two sides and an angle in a right-angled triangle.

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

The diagram shows the two sides we need for the sin ratio in purple: opposite and hypotenuse. The adjacent is shaded out because it is not needed for the sin ratio.

NOTE: These ratios are only valid when the triangle has a right-angle - otherwise the triangle would not have a hypotenuse. Always check that you have a right-angled triangle before you start using these trigonometric ratios. Otherwise you will not get the right answer.

Therefore, sin is defined as the opposite over the hypotenuse.

Submit your answer as: and

Writing trigonometric ratios

The diagram below shows a right-angled triangle, with sides $a$ , $b$ , and $c$ . The two acute angles are $A$ and $B$ . The diagram is not drawn to scale.

Determine the value of $\cos B$ for this triangle.

Answer:

\cos B =

$\cos B =$

expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides of the triangle. Remember that the opposite and adjacent depend on the position of the angle $B$ . Then use the definition of $\cos B$ to get the answer.

STEP: Evaluate

\cos B

$\cos B$ for the given triangle
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We have been given the lengths of the three sides of a right-angled triangle. We need to evaluate the ratio $\cos B$ . The definition of cosine is:

\begin{matrix} \cos B = \frac{adjacent}{hypotenuse} \end{matrix}

$\begin{array}{c}\cos B = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

We need to find the lengths for the adjacent and the hypotenuse. The adjacent side depends on the location of the angle $B$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the adjacent side. It is the side next to the angle $B$ .

In this case, the length of the adjacent is $c$ and the length of the hypotenuse is equal to $b$ . Therefore:

\begin{aligned} \cos B & = \frac{adjacent}{hypotenuse} \\ = \frac{c}{b} \end{aligned}

$\begin{aligned} \cos B & = \frac{\text{adjacent}}{\text{hypotenuse}} \\ & = \frac{c}{b} \end{aligned}$

Therefore, the answer is $\cos B = \dfrac{c}{b}$ .

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Trigonometric ratios: opposite and adjacent sides

Consider the diagram below:

The following equation is based on the triangle above.

\begin{aligned} \cos 50 ° = \frac{◻}{R} \end{aligned}

$\begin{aligned} \cos 50° = \frac{\Box}{R} \end{aligned}$

What is the quantity represented by $\Box$ ?

Answer:

The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by identifying the hypotenuse, adjacent, and opposite sides of the right-angled triangle given. Then use the following definition to answer the question:

\begin{matrix} \cos 50 ° = \frac{adjacent}{hypotenuse} \end{matrix}

$\begin{array}{c}\cos 50° = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

STEP: Use the definition of

\cos

$\cos$

50 °

$\text{50}\text{°}$ to get the answer
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We need to find the missing quantity in the equation:

\begin{aligned} \cos 50 ° = \frac{◻}{R} \end{aligned}

$\begin{aligned} \cos 50° = \frac{\Box}{R} \end{aligned}$

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

The side that does not touch the $\text{90}\text{°}$ angle is called the hypotenuse. In this question, the hypotenuse is the side $R$ . The names of the other two sides depends on the angle of interest. In this question, the angle of interest is the one whose size is $\text{50}\text{°}$ . Using this angle:

side $S$ is the adjacent side because it is next to (or adjacent to) the $\text{50}\text{°}$ angle.
side $Q$ is the opposite side because it is across from (or opposite to) the $\text{50}\text{°}$ angle.

In order to define the cosine of an acute angle in a right-angled triangle, we make use of the adjacent and hypotenuse sides. In this case:

\begin{matrix} \cos 50 ° = \frac{adjacent}{hypotenuse} \end{matrix}

$\begin{array}{c}\cos 50° = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

In the triangle diagram given, the label of the adjacent side is $S$ and the label for the hypotenuse side is $R$ . We have drawn circles around these labels in the diagram shown below:

NOTE: We have shaded the side

Q

$Q$ because it is not required in the definition of the cosine of the angle

50 °

$\text{50}\text{°}$ .

We can now use these labels to define cosine for the angle $\text{50}\text{°}$ :

\begin{aligned} \cos 50 ° = \frac{S}{R} \end{aligned}

$\begin{aligned}\cos 50° = \frac{S}{R}\end{aligned}$

Comparing this to the equation with the unknown quantity, we can see that the identity of $\Box$ is the quantity $S$ .

The missing quantity $\Box$ is $S$ .

Submit your answer as:

Trigonometric ratios

The following diagram shows a right-angled triangle. The sides are labelled with lengths $14$ , $16$ , and $0$ . The two acute angles are labelled $\alpha$ and $\beta$ . The diagram is not drawn to scale.

Determine the value of $\tan \beta$ in the following triangle.

INSTRUCTION: Give your answer in surd form, if necessary, and simplify your answer completely. Type sqrt( ) if you need to indicate a square root.

Answer:

\tan β =

$\tan \beta =$

expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides of the triangle. Then use the definition of $\tan \beta$ to get the answer.

STEP: Determine the value of

\tan β

$\tan \beta$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to determine the value of $\tan \beta$ .

To find $\tan \beta$ , we need to identify the opposite and the adjacent. The opposite side is across the triangle from the angle $\beta$ . The adjacent side is next to $\beta$ . The arrows and circles below show the those two sides. The opposite side is across the triangle from $\beta$ , and the adjacent side is next to $\beta$ .

In this case, the length of the opposite is $14$ and the length of the adjacent is equal to $0$ . Therefore:

\begin{aligned} \tan β & = \frac{14}{0} \\ = \frac{7}{0} \\ = \tilde{\infty} \end{aligned}

$\begin{aligned} \tan \beta & = \frac{14}{0} \\ & = \frac{7}{0} \\ & = \tilde{\infty} \end{aligned}$

Therefore, $\tan \beta = \tilde{\infty}$ .

Submit your answer as:

Trigonometric ratio definitions

The triangle $MNP$ below is right-angled at $M$ .

Use the triangle $MNP$ to answer the following questions:

What is the quantity represented by $\Box$ in the following equation?

$\begin{aligned} \cos \hat{N} = \frac{◻}{N P} \end{aligned}$ $\begin{aligned} \cos \hat{N} = \frac{\Box}{NP} \end{aligned}$
Answer:
The missing quantity is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the angle $\hat{N}$ to identify the hypotenuse, adjacent, and opposite sides of the triangle $MNP$ . After that, use the following definition to answer the question:

$\begin{matrix} \cos \hat{N} = \frac{adjacent}{hypotenuse} \end{matrix}$ $\begin{array}{c}\cos \hat{N} = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

STEP: Use the definition of $\cos\hat{N}$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

$\begin{aligned} \cos \hat{N} = \frac{◻}{N P} \end{aligned}$ $\begin{aligned} \cos \hat{N} = \frac{\Box}{NP} \end{aligned}$

The names of the sides of a right-angled triangle are hypotenuse, adjacent, and opposite sides. The side that does not touch the angle $\hat{M}$ is the hypotenuse. In this question, the hypotenuse is the side $NP$ . The names of the other two sides depend on the angle of interest. In this question, the angle of interest is angle $\hat{N}$ . Using this angle:
- side $MN$ is the adjacent side because it is next to (or adjacent to) the angle $\hat{N}$ .
- side $MP$ is the opposite side because it is across from (or opposite to) the angle $\hat{N}$ .
To define cosine we make use of the sides: adjacent and hypotenuse. The definition of cosine for the angle $\hat{N}$ is:

$\begin{matrix} \cos \hat{N} = \frac{adjacent}{hypotenuse} \end{matrix}$ $\begin{array}{c}\cos \hat{N} = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

From the triangle diagram, the label for the adjacent side is $MN$ , and the label for the hypotenuse side is $NP$ .

Comparing this to the equation with the unknown quantity, we can see that the identity of $\Box$ is the quantity $MN$ . We can now complete the equation:

$\begin{aligned} \cos \hat{N} = \frac{M N}{N P} \end{aligned}$ $\begin{aligned} \cos \hat{N} = \frac{MN}{NP} \end{aligned}$

NOTE:
We will redraw the triangle $MNP$ to show the two sides we need for the cosine of the angle $\hat{N}$ . These are: $MN$ and $NP$ .

The opposite side, $MP$ , has been shaded out because it is not needed for the cosine ratio.

The missing quantity is $MN$ .

Submit your answer as:
From the drop-down list below, choose the quantity that would make the following equation complete.

$\begin{aligned} ◻ \hat{P} = \frac{M N}{N P} \end{aligned}$ $\begin{aligned} \Box \; \hat{P} = \frac{MN}{NP} \end{aligned}$

Answer: The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The missing quantity is a trigonometric function. Using the angle $\hat{P}$ , identify the hypotenuse, adjacent, and opposite sides of the triangle $MNP$ . Then use any of the following definitions to answer the question:

$\begin{aligned} \cos \hat{P} & = \frac{adjacent}{hypotenuse} \\ \sin \hat{P} & = \frac{opposite}{hypotenuse} \\ \tan \hat{P} & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{array}{rl} \cos \hat{P} &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \hat{P} &= \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \tan \hat{P} &= \dfrac{\text{opposite}}{\text{adjacent}} \end{array}$

STEP: Identify the trigonometric ratio of the angle $\hat{P}$ equal to the ratio $\dfrac{MN}{NP}$
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

$\begin{aligned} ◻ \hat{P} = \frac{M N}{N P} \end{aligned}$ $\begin{aligned} \Box \; \hat{P} = \frac{MN}{NP} \end{aligned}$

The missing quantity is a trigonometric function. Trigonometric functions of acute angles of a right-angled triangle are equal to the ratio of two of the sides. The definitions of three trigonometric functions of the angle $\hat{P}$ using the names of the sides, are:

$\begin{aligned} \cos \hat{P} & = \frac{adjacent}{hypotenuse} \\ \sin \hat{P} & = \frac{opposite}{hypotenuse} \\ \tan \hat{P} & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{array}{rl} \cos \hat{P} &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \hat{P} &= \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \tan \hat{P} &= \dfrac{\text{opposite}}{\text{adjacent}} \end{array}$

We now need to identify the names of the sides: $MN$ and $NP$ . Then we will identify the missing trigonometric function.

For the angle $\hat{P}$ , the adjacent is the side $MP$ , the opposite is side $MN$ , and the hypotenuse is side $NP$ . The equation in the question statement has the ratio of the opposite side to the hypotenuse side. Using the equations of the definitions of the three trigonometric ratios above, we find that the missing quantity is $\sin$ . The completed equation is:

$\begin{aligned} \sin \hat{P} = \frac{M N}{N P} \end{aligned}$ $\begin{aligned} \sin \hat{P} = \frac{MN}{NP} \end{aligned}$

NOTE:
We will redraw the triangle $MNP$ to show the two sides we need for the sine of the angle $\hat{P}$ . These are: the opposite side $MN$ and the hypotenuse side $NP$ .

The missing quantity is $\sin$ .

Submit your answer as:

Evaluating trigonometric ratios

The diagram below shows a right-angled triangle with sides of length $18$ , $24$ , and $30$ . The two acute angles are labelled $x$ and $y$ .

Determine the value of $\sin y$ for this triangle.

INSTRUCTION: Give your answer as a simplified fraction.

Answer:

\sin y =

$\sin y =$

fraction

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the definition of the $\sin$ ratio. Then identify which sides of the triangle correspond to the sides in the definition to find the answer.

STEP: Determine the value of

\sin y

$\sin y$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to evaluate the ratio $\sin y$ .

The definition of $\sin y$ is

\begin{aligned} \sin y & = \frac{opposite}{hypotenuse} \end{aligned}

$\begin{aligned} \sin y & = \frac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$

So we need to identify the opposite and hypotenuse sides for the angle $y$ . The opposite side depends on the location of the angle $y$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the opposite side. It is the side across the triangle from the angle $y$ .

In this case, the length of the opposite is $\text{24}$ and the length of the hypotenuse is $\text{30}$ . Therefore:

\begin{aligned} \sin y & = \frac{opposite}{hypotenuse} \\ = \frac{24}{30} \\ = \frac{4}{5} \end{aligned}

$\begin{aligned} \sin y & = \frac{\text{opposite}}{\text{hypotenuse}} \\ & = \frac{24}{30} \\ & = \frac{4}{5} \end{aligned}$

Therefore, the answer is $\sin y = \dfrac{4}{5}$ .

Submit your answer as:

Exercises

Finding a trigonometric ratio

Answer the following questions about this right-angled triangle:

Complete the definition of $\tan \theta$ using the dropdown menus below:

Answer:
$\tan \theta$ is defined as the ratio of the and the .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Complete the definition of $\tan \theta$
[−2 points ⇒ 0 / 2 points left]

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[0.2cm] \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}$

For a right-angled triangle, the side across the triangle from the right-angle is called the "hypotenuse". For either of the other angles in the triangle, the side across the triangle from the angle is labelled as the "opposite". And the side next to this angle is called the "adjacent" side.

Therefore, $\tan \theta$ is defined as the ratio of the opposite and the adjacent.

Submit your answer as: and
What is the value of the ratio $\tan \theta$ for the given right-angled triangle?

Answer: $\tan \theta =$ .
numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify which two sides of the triangle correspond to the sides you used in the definition of $\tan \theta$ in Question 1, and take the ratio of these two sides.

STEP: Take the ratio of the correct two sides of the triangle
[−2 points ⇒ 0 / 2 points left]

This question asks us to determine $\tan \theta$ for the given right-angled triangle.

From Question 1, we know that $\tan \theta$ is defined as the ratio of the opposite and the adjacent. We need to identify which sides of the triangle are the opposite and adjacent, and take the ratio of their lengths.

For the given triangle, the opposite has length $\text{12}$ and the adjacent has length $\text{5}$ .

$\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ = \frac{12}{5} \end{aligned}$ $\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} \\[0.2cm] & = \frac{12}{5} \end{aligned}$

Therefore, $\tan \theta$ is $\dfrac{ 12 }{ 5 }$ .

Submit your answer as:

Finding a trigonometric ratio

Answer the following questions about this right-angled triangle:

Complete the definition of $\tan \theta$ using the dropdown menus below:

Answer:
$\tan \theta$ is defined as the over the .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Complete the definition of $\tan \theta$
[−2 points ⇒ 0 / 2 points left]

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[0.2cm] \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}$

For a right-angled triangle, the side across the triangle from the right-angle is called the "hypotenuse". For either of the other angles in the triangle, the side across the triangle from the angle is labelled as the "opposite". And the side next to this angle is called the "adjacent" side.

Therefore, $\tan \theta$ is defined as the opposite over the adjacent.

Submit your answer as: and
What is the value of the ratio $\tan \theta$ for the given right-angled triangle?

Answer: $\tan \theta =$ .
numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify which two sides of the triangle correspond to the sides you used in the definition of $\tan \theta$ in Question 1, and take the ratio of these two sides.

STEP: Take the ratio of the correct two sides of the triangle
[−2 points ⇒ 0 / 2 points left]

This question asks us to determine $\tan \theta$ for the given right-angled triangle.

From Question 1, we know that $\tan \theta$ is defined as the opposite over the adjacent. We need to identify which sides of the triangle are the opposite and adjacent, and take the ratio of their lengths.

For the given triangle, the opposite has length $\text{4}$ and the adjacent has length $\text{3}$ .

$\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ = \frac{4}{3} \end{aligned}$ $\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} \\[0.2cm] & = \frac{4}{3} \end{aligned}$

Therefore, $\tan \theta$ is $\dfrac{ 4 }{ 3 }$ .

Submit your answer as:

Finding a trigonometric ratio

Answer the following questions about this right-angled triangle:

Complete the definition of $\tan \theta$ using the dropdown menus below:

Answer:
$\tan \theta$ is defined as the over the .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Complete the definition of $\tan \theta$
[−2 points ⇒ 0 / 2 points left]

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[0.2cm] \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}$

For a right-angled triangle, the side across the triangle from the right-angle is called the "hypotenuse". For either of the other angles in the triangle, the side across the triangle from the angle is labelled as the "opposite". And the side next to this angle is called the "adjacent" side.

Therefore, $\tan \theta$ is defined as the opposite over the adjacent.

Submit your answer as: and
What is the value of the ratio $\tan \theta$ for the given right-angled triangle?

Answer: $\tan \theta =$ .
numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify which two sides of the triangle correspond to the sides you used in the definition of $\tan \theta$ in Question 1, and take the ratio of these two sides.

STEP: Take the ratio of the correct two sides of the triangle
[−2 points ⇒ 0 / 2 points left]

This question asks us to determine $\tan \theta$ for the given right-angled triangle.

From Question 1, we know that $\tan \theta$ is defined as the opposite over the adjacent. We need to identify which sides of the triangle are the opposite and adjacent, and take the ratio of their lengths.

For the given triangle, the opposite has length $\text{3}$ and the adjacent has length $\text{4}$ .

$\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ = \frac{3}{4} \end{aligned}$ $\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} \\[0.2cm] & = \frac{3}{4} \end{aligned}$

Therefore, $\tan \theta$ is $\dfrac{ 3 }{ 4 }$ .

Submit your answer as:

A connection between sine and cosine

The diagram below shows a right-angled triangle, with acute angles labelled $a$ and $b$ , $a \neq b$ . The sides of the triangle are labelled $A$ , $B$ , and $C$ .

Because this is a right-angled triangle, the following statement is true:

\begin{matrix} \sin a = \cos b \end{matrix}

$\begin{array}{c}\sin a = \cos b\end{array}$

Why is this statement true?

Answer:

$\sin a = \cos b$ because .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the sides of the triangle needed for the ratios $\sin a$ and $\cos b$ . What do these sides have in common?

STEP: Identify the sides used for

\sin a

$\sin a$ and

\cos b

$\cos b$
[−1 point ⇒ 0 / 1 points left]

The trigonometric ratios sine ( $\sin$ ) and cosine ( $\cos$ ) tell us about the ratio of different sides relative to a particular angle $\theta$ in a right-angled triangle.

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \end{aligned}$

For $\sin a$ , we need to identify the side opposite the angle $a$ . For $\cos b$ , we need to find the side adjacent to the angle $b$ . This is shown with the red and blue arrows below.

\begin{aligned} \sin a & = \frac{B}{C} \\ = \cos b \end{aligned}

$\begin{aligned} \sin a & = \frac{B}{C} \\ & = \cos b \end{aligned}$

This rule is always true for complementary angles (angles that add up to

90 °

$90°$ ) in a right-angled triangle. This is also why the cosine ratio is called cosine: it is the COmplementary function of SINE!

Therefore, the correct answer is $\sin a = \cos b$ because the side opposite $a$ is the side adjacent to $b$ .

Submit your answer as:

A connection between sine and cosine

The diagram below shows a right-angled triangle, with acute angles labelled $\phi$ and $\lambda$ , $\phi \neq \lambda$ . The sides of the triangle are labelled $R$ , $S$ , and $T$ .

Because this is a right-angled triangle, the following statement is true:

\begin{matrix} \sin λ = \cos ϕ \end{matrix}

$\begin{array}{c}\sin \lambda = \cos \phi\end{array}$

Why is this statement true?

Answer:

$\sin \lambda = \cos \phi$ because .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the sides of the triangle needed for the ratios $\sin \lambda$ and $\cos \phi$ . What do these sides have in common?

STEP: Identify the sides used for

\sin λ

$\sin \lambda$ and

\cos ϕ

$\cos \phi$
[−1 point ⇒ 0 / 1 points left]

The trigonometric ratios sine ( $\sin$ ) and cosine ( $\cos$ ) tell us about the ratio of different sides relative to a particular angle $\theta$ in a right-angled triangle.

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \end{aligned}$

For $\sin \lambda$ , we need to identify the side opposite the angle $\lambda$ . For $\cos \phi$ , we need to find the side adjacent to the angle $\phi$ . This is shown with the red and blue arrows below.

\begin{aligned} \sin λ & = \frac{T}{S} \\ = \cos ϕ \end{aligned}

$\begin{aligned} \sin \lambda & = \frac{T}{S} \\ & = \cos \phi \end{aligned}$

This rule is always true for complementary angles (angles that add up to

90 °

$90°$ ) in a right-angled triangle. This is also why the cosine ratio is called cosine: it is the COmplementary function of SINE!

Therefore, the correct answer is $\sin \lambda = \cos \phi$ because the side opposite $\lambda$ is the side adjacent to $\phi$ .

Submit your answer as:

A connection between sine and cosine

The diagram below shows a right-angled triangle, with acute angles labelled $a$ and $b$ , $a \neq b$ . The sides of the triangle are labelled $R$ , $S$ , and $T$ .

Because this is a right-angled triangle, the following statement is true:

\begin{matrix} \sin b = \cos a \end{matrix}

$\begin{array}{c}\sin b = \cos a\end{array}$

Why is this statement true?

Answer:

$\sin b = \cos a$ because .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the sides of the triangle needed for the ratios $\sin b$ and $\cos a$ . What do these sides have in common?

STEP: Identify the sides used for

\sin b

$\sin b$ and

\cos a

$\cos a$
[−1 point ⇒ 0 / 1 points left]

The trigonometric ratios sine ( $\sin$ ) and cosine ( $\cos$ ) tell us about the ratio of different sides relative to a particular angle $\theta$ in a right-angled triangle.

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}} \\[0.2cm] \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \end{aligned}$

For $\sin b$ , we need to identify the side opposite the angle $b$ . For $\cos a$ , we need to find the side adjacent to the angle $a$ . This is shown with the red and blue arrows below.

\begin{aligned} \sin b & = \frac{T}{S} \\ = \cos a \end{aligned}

$\begin{aligned} \sin b & = \frac{T}{S} \\ & = \cos a \end{aligned}$

This rule is always true for complementary angles (angles that add up to

90 °

$90°$ ) in a right-angled triangle. This is also why the cosine ratio is called cosine: it is the COmplementary function of SINE!

Therefore, the correct answer is $\sin b = \cos a$ because the side opposite $b$ is the side adjacent to $a$ .

Submit your answer as:

Opposite and adjacent sides: connecting to angles

The diagram below shows a right-angled triangle with sides $X$ , $Y$ , and $Z$ , and two acute angles: $\theta$ , and $\delta$ .

Which side is adjacent to the angle $\theta$ ?

Answer: The side adjacent to angle

θ

$\theta$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The side that is adjacent to the angle $\theta$ is situated next to the angle $\theta$ .

STEP: Identify the side adjacent to the angle

θ

$\theta$
[−1 point ⇒ 0 / 1 points left]

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

The side that does not touch the $\text{90}\text{°}$ angle is the hypotenuse. The hypotenuse is also the longest side of a right-angled triangle. In this question, the hypotenuse is the side $Z$ .

The side that touches the angle $\theta$ is the adjacent side.
The remaining side that is faraway from the angle $\theta$ is the opposite side.

There is another acute angle in the triangle, which we can call $\delta$ . If we change the angle of interest to $\delta$ , the labels opposite and adjacent will change. However, the hypotenuse remains the same.

We can summarise the above information using the diagram below:

NOTE: The sides

X

$X$ and

Y

$Y$ can each be the opposite side or the adjacent side. It depends on the angle of interest.

The side adjacent to angle $\theta$ is $X$ .

Submit your answer as:

Opposite and adjacent sides: connecting to angles

The diagram below shows a right-angled triangle with sides $X$ , $Y$ , and $Z$ , and two acute angles: $\alpha$ , and $\omega$ .

Which side is opposite to the angle $\alpha$ ?

Answer: The side opposite to angle

α

$\alpha$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The side that is opposite to the angle $\alpha$ is situated across from the angle $\alpha$ .

STEP: Identify the side opposite to the angle

α

$\alpha$
[−1 point ⇒ 0 / 1 points left]

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

The side that does not touch the $\text{90}\text{°}$ angle is the hypotenuse. The hypotenuse is also the longest side of a right-angled triangle. In this question, the hypotenuse is the side $Y$ .

We label the other two sides according to the location of the angle of interest. This question asked about the sides adjacent and opposite to the angle $\alpha$ , so the angle of interest is angle $\alpha$ . Using this angle, we label the other two sides in this manner:

The side that touches the angle $\alpha$ is the adjacent side.
The remaining side that is faraway from the angle $\alpha$ is the opposite side.

We can summarise the above information using the diagram below:

NOTE: The sides

X

$X$ and

Z

$Z$ can each be the opposite side or the adjacent side. It depends on the angle of interest.

The side opposite to angle $\alpha$ is $Z$ .

Submit your answer as:

Opposite and adjacent sides: connecting to angles

The diagram below shows a right-angled triangle with sides $Q$ , $R$ , and $S$ , and two acute angles: $\beta$ , and $\alpha$ .

Which side is opposite to the angle $\alpha$ ?

Answer: The side opposite to angle

α

$\alpha$ is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The side that is opposite to the angle $\alpha$ is situated across from the angle $\alpha$ .

STEP: Identify the side opposite to the angle

α

$\alpha$
[−1 point ⇒ 0 / 1 points left]

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

The side that does not touch the $\text{90}\text{°}$ angle is the hypotenuse. The hypotenuse is also the longest side of a right-angled triangle. In this question, the hypotenuse is the side $S$ .

The side that touches the angle $\alpha$ is the adjacent side.
The remaining side that is faraway from the angle $\alpha$ is the opposite side.

There is another acute angle in the triangle, which we can call $\beta$ . If we change the angle of interest to $\beta$ , the labels opposite and adjacent will change. However, the hypotenuse remains the same.

We can summarise the above information using the diagram below:

NOTE: The sides

R

$R$ and

Q

$Q$ can each be the opposite side or the adjacent side. It depends on the angle of interest.

The side opposite to angle $\alpha$ is $Q$ .

Submit your answer as:

Working with trigonometric ratios

Determine the value of $\tan \theta$ in the following triangle.

Select an answer from the following list:

A	$\dfrac{\sqrt{3}}{2}$
B	$\dfrac{\sqrt{2}}{2}$
C	$\dfrac{1}{2}$
D	$1$
E	$\dfrac{\sqrt{3}}{3}$
F	$\sqrt{3}$

Answer:

\tan θ =

$\tan \theta =$

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the opposite, adjacent, and hypotenuse sides of the triangle. Then use the definition of $\tan \theta$ to get the answer.

STEP: Use the definition of

\tan θ

$\tan \theta$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to determine the value of $\tan \theta$ . The definition of the tangent ratio is:

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \end{aligned}

$\begin{aligned} \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}$

In this case the length of the opposite side is $5 \sqrt{2}$ and the length of the adjacent side is $5 \sqrt{2}$ . Therefore,

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ = \frac{5 \sqrt{2}}{5 \sqrt{2}} \\ = 1 \end{aligned}

$\begin{aligned} \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \\ & = \dfrac{5 \sqrt{2}}{5 \sqrt{2}} \\ & = 1 \end{aligned}$

NOTE: The hypotenuse of the triangle is greyed out in the diagram above. That is because it is not needed for this question: the tangent ratio does not care about that side of the triangle.

The correct choice from the list is D.

Submit your answer as:

Working with trigonometric ratios

Determine the value of $\tan \theta$ in the following triangle.

Select an answer from the following list:

A	$\dfrac{\sqrt{3}}{2}$
B	$\dfrac{\sqrt{2}}{2}$
C	$\dfrac{1}{2}$
D	$1$
E	$\dfrac{\sqrt{3}}{3}$
F	$\sqrt{3}$

Answer:

\tan θ =

$\tan \theta =$

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the opposite, adjacent, and hypotenuse sides of the triangle. Then use the definition of $\tan \theta$ to get the answer.

STEP: Use the definition of

\tan θ

$\tan \theta$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to determine the value of $\tan \theta$ . The definition of the tangent ratio is:

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \end{aligned}

$\begin{aligned} \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \end{aligned}$

In this case the length of the opposite side is $6 \sqrt{3}$ and the length of the adjacent side is $18$ . Therefore,

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ = \frac{6 \sqrt{3}}{18} \\ = \frac{\sqrt{3}}{3} \end{aligned}

$\begin{aligned} \tan \theta & = \dfrac{\text{opposite}}{\text{adjacent}} \\ & = \dfrac{6 \sqrt{3}}{18} \\ & = \frac{\sqrt{3}}{3} \end{aligned}$

NOTE: The hypotenuse of the triangle is greyed out in the diagram above. That is because it is not needed for this question: the tangent ratio does not care about that side of the triangle.

The correct choice from the list is E.

Submit your answer as:

Working with trigonometric ratios

Determine the value of $\sin \theta$ in the following triangle.

Select an answer from the following list:

A	$\dfrac{\sqrt{3}}{2}$
B	$\dfrac{\sqrt{2}}{2}$
C	$\dfrac{1}{2}$
D	$1$
E	$\dfrac{\sqrt{3}}{3}$
F	$\sqrt{3}$

Answer:

\sin θ =

$\sin \theta =$

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the opposite, adjacent, and hypotenuse sides of the triangle. Then use the definition of $\sin \theta$ to get the answer.

STEP: Use the definition of

\sin θ

$\sin \theta$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to determine the value of $\sin \theta$ . The definition of the sine ratio is:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \end{aligned}

$\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$

We need to find the lengths for the opposite side and the hypotenuse. The opposite side depends on the location of angle $\theta$ . (The hypotenuse is always the side of the triangle across from the right angle.) The arrow and circle below show the opposite side. It is the side across from the angle $\theta$ .

In this case the length of the opposite side is $6 \sqrt{2}$ and the length of the hypotenuse is $12$ . Therefore,

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ = \frac{6 \sqrt{2}}{12} \\ = \frac{\sqrt{2}}{2} \end{aligned}

$\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\ & = \dfrac{6 \sqrt{2}}{12} \\ & = \frac{\sqrt{2}}{2} \end{aligned}$

NOTE: The adjacent side of the triangle is greyed out in the diagram above. That is because it is not needed for this question: the sine ratio does not care about that side of the triangle.

The correct choice from the list is B.

Submit your answer as:

Right triangles: opposite and adjacent sides

The diagram below shows a right-angled triangle with sides labelled Side $\text{1}$ , Side $\text{2}$ , and Side $\text{3}$ . There is also an acute angle $\phi$ .

Which sides are adjacent and opposite to the angle $\phi$ ?

Answer:

The side adjacent to angle $\phi$ is .

The side opposite to angle $\phi$ is .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The side that is adjacent to the angle $\phi$ is next to the angle $\phi$ . The side that is opposite to the angle $\phi$ is across from the angle $\phi$ .

STEP: Identify the sides opposite and adjacent to the angle

ϕ

$\phi$
[−2 points ⇒ 0 / 2 points left]

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

We label the other two sides according to the location of the angle of interest. This question asked about the sides adjacent and opposite to the angle $\phi$ , so the angle of interest is angle $\phi$ . Using this angle, we label the other two sides in this manner:

The side that touches the angle $\phi$ is the adjacent side.
The remaining side that is far away from the angle $\phi$ is the opposite side.

There is another acute angle in the triangle, which we can call $\alpha$ . If we change the angle of interest to $\alpha$ , the labels opposite and adjacent will change. However, the hypotenuse remains the same.

We can summarise the above information using the diagram below:

NOTE: Side

2

$\text{2}$ and Side

1

$\text{1}$ can each be the opposite side or the adjacent side. It depends on the angle of interest.

The side adjacent to angle $\phi$ is Side $\text{2}$ .

The side opposite to angle $\phi$ is Side $\text{1}$ .

Submit your answer as: and

Right triangles: opposite and adjacent sides

The diagram below shows a right-angled triangle with sides labelled Side $\text{1}$ , Side $\text{2}$ , and Side $\text{3}$ . There is also an acute angle $\omega$ .

Which sides are adjacent and opposite to the angle $\omega$ ?

Answer:

The side adjacent to angle $\omega$ is .

The side opposite to angle $\omega$ is .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The side that is adjacent to the angle $\omega$ is next to the angle $\omega$ . The side that is opposite to the angle $\omega$ is across from the angle $\omega$ .

STEP: Identify the sides opposite and adjacent to the angle

ω

$\omega$
[−2 points ⇒ 0 / 2 points left]

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

We label the other two sides according to the location of the angle of interest. This question asked about the sides adjacent and opposite to the angle $\omega$ , so the angle of interest is angle $\omega$ . Using this angle, we label the other two sides in this manner:

The side that touches the angle $\omega$ is the adjacent side.
The remaining side that is far away from the angle $\omega$ is the opposite side.

We can summarise the above information using the diagram below:

NOTE: Side

2

$\text{2}$ and Side

1

$\text{1}$ can each be the opposite side or the adjacent side. It depends on the angle of interest.

The side adjacent to angle $\omega$ is Side $\text{2}$ .

The side opposite to angle $\omega$ is Side $\text{1}$ .

Submit your answer as: and

Right triangles: opposite and adjacent sides

The diagram below shows a right-angled triangle with sides labelled Side $\text{1}$ , Side $\text{2}$ , and Side $\text{3}$ . There is also an acute angle $\theta$ .

Which sides are adjacent and opposite to the angle $\theta$ ?

Answer:

The side adjacent to angle $\theta$ is .

The side opposite to angle $\theta$ is .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The side that is adjacent to the angle $\theta$ is next to the angle $\theta$ . The side that is opposite to the angle $\theta$ is across from the angle $\theta$ .

STEP: Identify the sides opposite and adjacent to the angle

θ

$\theta$
[−2 points ⇒ 0 / 2 points left]

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

The side that touches the angle $\theta$ is the adjacent side.
The remaining side that is far away from the angle $\theta$ is the opposite side.

We can summarise the above information using the diagram below:

NOTE: Side

3

$\text{3}$ and Side

1

$\text{1}$ can each be the opposite side or the adjacent side. It depends on the angle of interest.

The side adjacent to angle $\theta$ is Side $\text{3}$ .

The side opposite to angle $\theta$ is Side $\text{1}$ .

Submit your answer as: and

Calculation of the area of a triangle

A triangle $ABC$ is shown in the figure below. The length of $\overline{AC}$ is $\text{4,4}$ units. The size of the angle $B\hat{A}C$ is $\text{34}°$ .

Work out the lengths of the line segments $\overline{AB}$ and $\overline{BC}$ . You will need to use your knowledge of trigonometric ratios for these calculations. Round your answers to 2 decimal places.
Answer:
$\overline{AB} =$ units
$\overline{BC} =$ units

numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Use the definitions of the sine and cosine functions.

STEP: Choose the correct trigonometric ratios
[−2 points ⇒ 2 / 4 points left]

We are given the following information:
- The triangle has a right angle ( $A\hat{B}C$ ) (shown in the diagram)
- $\overline{AC} = \text{4,4}$ units
- $B\hat{A}C = \text{34}°$
This means we can use the cosine ( $\cos$ ) and sine ( $\sin$ ) functions to detemine the lengths of the line segments $\overline{AB}$ and $\overline{BC}$ . The $\cos$ and $\sin$ of an angle, for example, $\theta$ are defined as follows:

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \sin θ & = \frac{opposite}{hypotenuse} \end{aligned}$ $\begin{aligned} \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$

Line segment $\overline{AB}$ is adjacent to the angle $B\hat{A}C$ and line segment $\overline{BC}$ is opposite to the angle $B\hat{A}C$ . Line segment $\overline{AC}$ is the hypotenuse (it is opposite the right angle $A\hat{B}C$ ). We know the length of $\overline{AC}$ and the size of angle $B\hat{A}C$ , so we can calculate the unkown line segments.

Therefore, we will use the cosine function to calculate the length of $\overline{AB}$ and the sine function to calculate the length of $\overline{BC}$ .

STEP: Use the trigonometric ratios to find the sides of the triangle
[−2 points ⇒ 0 / 4 points left]

We will now calculate the lengths of $\overline{AB}$ and $\overline{BC}$ .

For $\overline{AB}$ :

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (B \hat{A} C) & = \frac{\bar{A B}}{\bar{A C}} \\ ∴ \bar{A B} & = \cos (B \hat{A} C) \times \bar{A C} \\ = \cos 34 ° \times 4,4 \\ = 3,64776... \\ \approx 3,65 units \end{aligned}$ $\begin{aligned} \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \cos (B\hat{A}C) & = \dfrac{\phantom{=} \overline{AB} \phantom{=}}{\overline{AC}} \\ \therefore \overline{AB} & = \cos (B\hat{A}C) \times \overline{AC} \\ & = \cos \text{34}° \times \text{4,4} \\ & = \text{3,64776...} \\ & \approx \text{3,65} \text{ units} \end{aligned}$

For $\overline{BC}$ :

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin (B \hat{A} C) & = \frac{\bar{B C}}{\bar{A C}} \\ ∴ \bar{B C} & = \sin (B \hat{A} C) \times \bar{A C} \\ = \sin 34 ° \times 4,4 \\ = 2,46044... \\ \approx 2,46 units \end{aligned}$ $\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \sin (B\hat{A}C) & = \dfrac{\phantom{=} \overline{BC} \phantom{=}}{\overline{AC}} \\ \therefore \overline{BC} & = \sin (B\hat{A}C) \times \overline{AC} \\ & = \sin \text{34}° \times \text{4,4} \\ & = \text{2,46044...} \\ & \approx \text{2,46} \text{ units} \end{aligned}$

Therefore, the length of $\overline{AB} = \text{3,65}$ units and the length of $\overline{BC} = \text{2,46}$ units.

Submit your answer as: and

Which formula from the table below would you use to calculate the area of the triangle $ABC$ ?

Choice	Formula
A	$\frac{1}{2} \times \overline{AB} \times \overline{BC} \times \overline{CA}$
B	$\overline{AB} \times \overline{BC} \times \overline{CA}$
C	$\overline{AB} \times \overline{BC}$
D	$\pi \times \left(\overline{AB}\right)^2$
E	$\frac{1}{2} \times \overline{AB} \times \overline{BC}$

Answer: The correct formula is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Consider a right triangle to be a result of dividing a rectangle into two parts along the diagonal. Do you remember the formula for the area of a rectangle?

STEP: Determine the formula using the relationship between a right triangle and a rectangle
[−1 point ⇒ 0 / 1 points left]

We can consider the triangle $ABC$ part of a rectangle $ABCD$ , as shown below:

You can determine the area of any right angled triangle by multiplying the two sides adjacent to the right angle, and dividing by 2 (or multiplying by a half).

Therefore, the correct formula is choice E.

Submit your answer as:

Using the unrounded answers from the first question, calculate the area of the triangle $ABC$ . Round your answers to 2 decimal places.

Answer: Area of triangle $ABC =$ $\text{units}^2$
numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the area of a triangle.

STEP: Use the formula for the area of a triangle
[−2 points ⇒ 0 / 2 points left]

From the previous question, we have found that the formula for the area of the triangle $ABC$ is $\frac{1}{2} \times \overline{AB} \times \overline{BC}$ .

From Question 1, we found that $\overline{AB} = \text{3,64776...}$ units and $\overline{BC} = \text{2,46044...}$ units.

The area of the triangle $ABC$ :

$\begin{aligned} Area of triangle A B C & = \frac{1}{2} \times \bar{A B} \times \bar{B C} \\ = \frac{1}{2} \times 3,64776... \times 2,46044... \\ = 4,48756... \\ \approx 4,49 {units}^{2} \end{aligned}$ $\begin{aligned} \text{Area of triangle } ABC & = \frac{1}{2} \times \overline{AB} \times \overline{BC} \\ & = \frac{1}{2} \times \text{3,64776...} \times \text{2,46044...} \\ & = \text{4,48756...} \\ & \approx \text{4,49} \text{ units}^2 \end{aligned}$

Therefore, the area of the triangle $ABC$ is $\text{4,49}$ units $^2$ .

Submit your answer as:

Calculation of the area of a triangle

A triangle $ABC$ is shown in the figure below. The length of $\overline{AC}$ is $\text{4,2}$ units. The size of the angle $B\hat{A}C$ is $\text{39}°$ .

Calculate the lengths of the line segments $\overline{AB}$ and $\overline{BC}$ . You will need to use your knowledge of trigonometric ratios for these calculations. Round your answers to 2 decimal places.
Answer:
$\overline{AB} =$ units
$\overline{BC} =$ units

numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Use the definitions of the sine and cosine functions.

STEP: Choose the correct trigonometric ratios
[−2 points ⇒ 2 / 4 points left]

We are given the following information:
- The triangle has a right angle ( $A\hat{B}C$ ) (shown in the diagram)
- $\overline{AC} = \text{4,2}$ units
- $B\hat{A}C = \text{39}°$
This means we can use the cosine ( $\cos$ ) and sine ( $\sin$ ) functions to detemine the lengths of the line segments $\overline{AB}$ and $\overline{BC}$ . The $\cos$ and $\sin$ of an angle, for example, $\theta$ are defined as follows:

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \sin θ & = \frac{opposite}{hypotenuse} \end{aligned}$ $\begin{aligned} \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$

Line segment $\overline{AB}$ is adjacent to the angle $B\hat{A}C$ and line segment $\overline{BC}$ is opposite to the angle $B\hat{A}C$ . Line segment $\overline{AC}$ is the hypotenuse (it is opposite the right angle $A\hat{B}C$ ). We know the length of $\overline{AC}$ and the size of angle $B\hat{A}C$ , so we can calculate the unkown line segments.

Therefore, we will use the cosine function to calculate the length of $\overline{AB}$ and the sine function to calculate the length of $\overline{BC}$ .

STEP: Use the trigonometric ratios to find the sides of the triangle
[−2 points ⇒ 0 / 4 points left]

We will now calculate the lengths of $\overline{AB}$ and $\overline{BC}$ .

For $\overline{AB}$ :

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (B \hat{A} C) & = \frac{\bar{A B}}{\bar{A C}} \\ ∴ \bar{A B} & = \cos (B \hat{A} C) \times \bar{A C} \\ = \cos 39 ° \times 4,2 \\ = 3,26401... \\ \approx 3,26 units \end{aligned}$ $\begin{aligned} \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \cos (B\hat{A}C) & = \dfrac{\phantom{=} \overline{AB} \phantom{=}}{\overline{AC}} \\ \therefore \overline{AB} & = \cos (B\hat{A}C) \times \overline{AC} \\ & = \cos \text{39}° \times \text{4,2} \\ & = \text{3,26401...} \\ & \approx \text{3,26} \text{ units} \end{aligned}$

For $\overline{BC}$ :

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin (B \hat{A} C) & = \frac{\bar{B C}}{\bar{A C}} \\ ∴ \bar{B C} & = \sin (B \hat{A} C) \times \bar{A C} \\ = \sin 39 ° \times 4,2 \\ = 2,64314... \\ \approx 2,64 units \end{aligned}$ $\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \sin (B\hat{A}C) & = \dfrac{\phantom{=} \overline{BC} \phantom{=}}{\overline{AC}} \\ \therefore \overline{BC} & = \sin (B\hat{A}C) \times \overline{AC} \\ & = \sin \text{39}° \times \text{4,2} \\ & = \text{2,64314...} \\ & \approx \text{2,64} \text{ units} \end{aligned}$

Therefore, the length of $\overline{AB} = \text{3,26}$ units and the length of $\overline{BC} = \text{2,64}$ units.

Submit your answer as: and

Which formula from the table below would you use to calculate the area of the triangle $ABC$ ?

Choice	Formula
A	$\overline{AB} \times \overline{BC}$
B	$\frac{1}{2} \times \overline{AB} \times \overline{BC}$
C	$\overline{AB} \times \overline{BC} \times \overline{CA}$
D	$\frac{1}{2} \times \overline{AB} \times \overline{BC} \times \overline{CA}$
E	$\pi \times \left(\overline{AB}\right)^2$

Answer: The correct formula is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Consider a right triangle to be a result of dividing a rectangle into two parts along the diagonal. Do you remember the formula for the area of a rectangle?

STEP: Determine the formula using the relationship between a right triangle and a rectangle
[−1 point ⇒ 0 / 1 points left]

We can consider the triangle $ABC$ part of a rectangle $ABCD$ , as shown below:

You can determine the area of any right angled triangle by multiplying the two sides adjacent to the right angle, and dividing by 2 (or multiplying by a half).

Therefore, the correct formula is choice B.

Submit your answer as:

Using the unrounded answers from the first question, work out the area of the triangle $ABC$ . Round your answers to 2 decimal places.

Answer: Area of triangle $ABC =$ $\text{units}^2$
numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the area of a triangle.

STEP: Use the formula for the area of a triangle
[−2 points ⇒ 0 / 2 points left]

From the previous question, we have found that the formula for the area of the triangle $ABC$ is $\frac{1}{2} \times \overline{AB} \times \overline{BC}$ .

From Question 1, we found that $\overline{AB} = \text{3,26401...}$ units and $\overline{BC} = \text{2,64314...}$ units.

The area of the triangle $ABC$ :

$\begin{aligned} Area of triangle A B C & = \frac{1}{2} \times \bar{A B} \times \bar{B C} \\ = \frac{1}{2} \times 3,26401... \times 2,64314... \\ = 4,31363... \\ \approx 4,31 {units}^{2} \end{aligned}$ $\begin{aligned} \text{Area of triangle } ABC & = \frac{1}{2} \times \overline{AB} \times \overline{BC} \\ & = \frac{1}{2} \times \text{3,26401...} \times \text{2,64314...} \\ & = \text{4,31363...} \\ & \approx \text{4,31} \text{ units}^2 \end{aligned}$

Therefore, the area of the triangle $ABC$ is $\text{4,31}$ units $^2$ .

Submit your answer as:

Calculation of the area of a triangle

A triangle $ABC$ is shown in the figure below. The length of $\overline{AC}$ is $\text{5}$ units. The size of the angle $B\hat{A}C$ is $\text{35}°$ .

Calculate the lengths of the line segments $\overline{AB}$ and $\overline{BC}$ . You will need to use your knowledge of trigonometric ratios for these calculations. Round your answers to 2 decimal places.
Answer:
$\overline{AB} =$ units
$\overline{BC} =$ units

numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Use the definitions of the sine and cosine functions.

STEP: Choose the correct trigonometric ratios
[−2 points ⇒ 2 / 4 points left]

We are given the following information:
- The triangle has a right angle ( $A\hat{B}C$ ) (shown in the diagram)
- $\overline{AC} = \text{5}$ units
- $B\hat{A}C = \text{35}°$
This means we can use the cosine ( $\cos$ ) and sine ( $\sin$ ) functions to detemine the lengths of the line segments $\overline{AB}$ and $\overline{BC}$ . The $\cos$ and $\sin$ of an angle, for example, $\theta$ are defined as follows:

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \sin θ & = \frac{opposite}{hypotenuse} \end{aligned}$ $\begin{aligned} \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$

Line segment $\overline{AB}$ is adjacent to the angle $B\hat{A}C$ and line segment $\overline{BC}$ is opposite to the angle $B\hat{A}C$ . Line segment $\overline{AC}$ is the hypotenuse (it is opposite the right angle $A\hat{B}C$ ). We know the length of $\overline{AC}$ and the size of angle $B\hat{A}C$ , so we can calculate the unkown line segments.

Therefore, we will use the cosine function to calculate the length of $\overline{AB}$ and the sine function to calculate the length of $\overline{BC}$ .

STEP: Use the trigonometric ratios to find the sides of the triangle
[−2 points ⇒ 0 / 4 points left]

We will now calculate the lengths of $\overline{AB}$ and $\overline{BC}$ .

For $\overline{AB}$ :

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (B \hat{A} C) & = \frac{\bar{A B}}{\bar{A C}} \\ ∴ \bar{A B} & = \cos (B \hat{A} C) \times \bar{A C} \\ = \cos 35 ° \times 5 \\ = 4,09576... \\ \approx 4,1 units \end{aligned}$ $\begin{aligned} \cos \theta & = \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \cos (B\hat{A}C) & = \dfrac{\phantom{=} \overline{AB} \phantom{=}}{\overline{AC}} \\ \therefore \overline{AB} & = \cos (B\hat{A}C) \times \overline{AC} \\ & = \cos \text{35}° \times \text{5} \\ & = \text{4,09576...} \\ & \approx \text{4,1} \text{ units} \end{aligned}$

For $\overline{BC}$ :

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin (B \hat{A} C) & = \frac{\bar{B C}}{\bar{A C}} \\ ∴ \bar{B C} & = \sin (B \hat{A} C) \times \bar{A C} \\ = \sin 35 ° \times 5 \\ = 2,86788... \\ \approx 2,87 units \end{aligned}$ $\begin{aligned} \sin \theta & = \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \sin (B\hat{A}C) & = \dfrac{\phantom{=} \overline{BC} \phantom{=}}{\overline{AC}} \\ \therefore \overline{BC} & = \sin (B\hat{A}C) \times \overline{AC} \\ & = \sin \text{35}° \times \text{5} \\ & = \text{2,86788...} \\ & \approx \text{2,87} \text{ units} \end{aligned}$

Therefore, the length of $\overline{AB} = \text{4,1}$ units and the length of $\overline{BC} = \text{2,87}$ units.

Submit your answer as: and

Which formula from the table below would you use to calculate the area of the triangle $ABC$ ?

Choice	Formula
A	$\overline{AB} \times \overline{BC}$
B	$\frac{1}{2} \times \overline{AB} \times \overline{BC}$
C	$\frac{1}{2} \times \overline{AB} \times \overline{BC} \times \overline{CA}$
D	$\overline{AB} \times \overline{BC} \times \overline{CA}$
E	$\pi \times \left(\overline{AB}\right)^2$

Answer: The correct formula is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Consider a right triangle to be a result of dividing a rectangle into two parts along the diagonal. Do you remember the formula for the area of a rectangle?

STEP: Determine the formula using the relationship between a right triangle and a rectangle
[−1 point ⇒ 0 / 1 points left]

We can consider the triangle $ABC$ part of a rectangle $ABCD$ , as shown below:

You can determine the area of any right angled triangle by multiplying the two sides adjacent to the right angle, and dividing by 2 (or multiplying by a half).

Therefore, the correct formula is choice B.

Submit your answer as:

Using the unrounded answers from the first question, determine the area of the triangle $ABC$ . Round your answers to 2 decimal places.

Answer: Area of triangle $ABC =$ $\text{units}^2$
numeric

HINT: <no title>
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Use the formula for the area of a triangle.

STEP: Use the formula for the area of a triangle
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From the previous question, we have found that the formula for the area of the triangle $ABC$ is $\frac{1}{2} \times \overline{AB} \times \overline{BC}$ .

From Question 1, we found that $\overline{AB} = \text{4,09576...}$ units and $\overline{BC} = \text{2,86788...}$ units.

The area of the triangle $ABC$ :

$\begin{aligned} Area of triangle A B C & = \frac{1}{2} \times \bar{A B} \times \bar{B C} \\ = \frac{1}{2} \times 4,09576... \times 2,86788... \\ = 5,87307... \\ \approx 5,87 {units}^{2} \end{aligned}$ $\begin{aligned} \text{Area of triangle } ABC & = \frac{1}{2} \times \overline{AB} \times \overline{BC} \\ & = \frac{1}{2} \times \text{4,09576...} \times \text{2,86788...} \\ & = \text{5,87307...} \\ & \approx \text{5,87} \text{ units}^2 \end{aligned}$

Therefore, the area of the triangle $ABC$ is $\text{5,87}$ units $^2$ .

Submit your answer as:

Identify and use the correct trigonometric ratio

The diagram below shows right-angled triangle $\Delta NPQ$ with $\angle N = \text{90} \text{°}$ . The sides given are: $\overline{ NP } = \text{6,93}$ and $\overline{ PQ } = \text{8}$ .

Determine the value of $\hat{NPQ}$ . Round your answer to two decimal places if appropriate.

Answer:

∠ N P Q =

$\angle NPQ =$

°

${°}$

numeric

HINT: <no title>
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The first thing you need to do is figure out which trigonometric ratio to use: sine, cosine, or tangent.

STEP: Find the correct trigonometric ratio to solve for the unkown
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We need to determine the size of angle $\angle NPQ$ in the right-triangle, $\Delta NPQ$ . We need to determine which trigonometric ratio we can use to find $\angle NPQ$ . Side $\overline{ NP }$ is adjacent to the unknown angle. And side $\overline{ PQ }$ is the hypotenuse. The trigonometric ratio linking the hypotenuse to the adjacent side is the cosine ratio. So we can set up an equation with this information like below, using $\theta$ to represent the unknown angle.

\begin{aligned} \cos θ & = \frac{\bar{N P}}{\bar{P Q}} \end{aligned}

$\begin{aligned} \cos\theta &= \frac{\overline{ NP }}{\overline{ PQ }} \end{aligned}$

STEP: Solve the trigonometric equation
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Substitute the given values into the equation. Then solve the equation.

\begin{aligned} \cos θ & = \frac{6,93}{8} \\ θ & = \cos^{- 1} (\frac{6,93}{8}) \\ ∠ N P Q s & = 29,97425... ° \\ ∠ N P Q & \approx 29,97 ° \end{aligned}

$\begin{aligned} \cos\theta &= \frac{\text{6,93}}{\text{8}}\\ \theta &= \cos^{-1} \left(\frac{\text{6,93}}{\text{8}}\right)\\ \angle NPQ s&= \text{29,97425...} \text{°}\\ \angle NPQ & \approx \text{29,97}\text{°} \end{aligned}$

Remember that the instructions say to round the answer to the second decimal place, as shown in the last step above.

The size of $\angle NPQ$ is $\text{29,97}\text{°}$ .

Submit your answer as:

Identify and use the correct trigonometric ratio

The diagram below shows right-angled triangle $\Delta ABC$ with $\angle A = \text{90} \text{°}$ . The sides given are: $\overline{ AB } = \text{4,95}$ and $\overline{ BC } = \text{7}$ .

Determine the value of $\angle ABC$ . Round your answer to two decimal places if appropriate.

Answer:

∠ A B C =

$\angle ABC =$

°

${°}$

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The first thing you need to do is figure out which trigonometric ratio to use: sine, cosine, or tangent.

STEP: Find the correct trigonometric ratio to solve for the unkown
[−2 points ⇒ 1 / 3 points left]

We need to determine the size of angle $\angle ABC$ in the right-triangle, $\Delta ABC$ . We need to determine which trigonometric ratio we can use to find $\angle ABC$ . Side $\overline{ AB }$ is adjacent to the unknown angle. And side $\overline{ BC }$ is the hypotenuse. The trigonometric ratio linking the hypotenuse to the adjacent side is the cosine ratio. So we can set up an equation with this information like below, using $\alpha$ to represent the unknown angle.

\begin{aligned} \cos α & = \frac{\bar{A B}}{\bar{B C}} \end{aligned}

$\begin{aligned} \cos\alpha &= \frac{\overline{ AB }}{\overline{ BC }} \end{aligned}$

STEP: Solve the trigonometric equation
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Substitute the given values into the equation. Then solve the equation.

\begin{aligned} \cos α & = \frac{4,95}{7} \\ α & = \cos^{- 1} (\frac{4,95}{7}) \\ ∠ A B C s & = 44,99707... ° \\ ∠ A B C & \approx 45,0 ° \end{aligned}

$\begin{aligned} \cos\alpha &= \frac{\text{4,95}}{\text{7}}\\ \alpha &= \cos^{-1} \left(\frac{\text{4,95}}{\text{7}}\right)\\ \angle ABC s&= \text{44,99707...} \text{°}\\ \angle ABC & \approx \text{45,0}\text{°} \end{aligned}$

Remember that the instructions say to round the answer to the second decimal place, as shown in the last step above.

The size of $\angle ABC$ is $\text{45,0}\text{°}$ .

Submit your answer as:

Identify and use the correct trigonometric ratio

Study the given right-angled triangle $\Delta XGV$ with $\angle X = \text{90} \text{°}$ and answer the following question. The angle given in the diagram is: $\angle GVX = \text{51,34}\text{°}$ . The side given is: $\overline{ GX } = \text{5}$ .

Determine the length of side $\overline{XV}$ . Round your answer to two decimal places if appropriate.

Answer:

\bar{X V} =

$\overline{ XV } =$

numeric

HINT: <no title>
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The first thing you need to do is figure out which trigonometric ratio to use: sine, cosine, or tangent.

STEP: Find the correct trigonometric ratio to solve for the unkown
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The question gives us the side $\overline{ GX } = \text{5}$ and $\angle GVX = \text{51,34}°$ . And we need to find side $\overline{ XV }$ . Side $\overline{ GX }$ is opposite the given angle and side $\overline{ XV }$ is adjacent to the angle. This combination corresponds to the tangent ratio. Set up an equation with this information and call the missing side $y$ .

\begin{aligned} \tan 51,34 ° & = \frac{\bar{G X}}{\bar{X V}} \\ \tan 51,34 ° & = \frac{\bar{G X}}{y} \end{aligned}

$\begin{aligned} \tan \text{51,34}{°} &= \frac{\overline{ GX }}{\overline{ XV }} \\ \tan \text{51,34}{°} &= \frac{\overline{ GX }}{y} \end{aligned}$

STEP: Solve the trigonometric equation
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Substitute the given values into the equation we set up in the first step, and solve the equation.

\begin{aligned} \tan 51,34 ° & = \frac{5}{y} \\ y \cdot \tan 51,34 ° & = 5 \\ y & = \frac{5}{\tan 51,34 °} \\ y & = 4,00002... \\ y & \approx 4 \end{aligned}

$\begin{aligned} \tan \text{51,34}° &= \frac{\text{5}}{y}\\ y \cdot \tan \text{51,34}° &= \text{5}\\ y &= \frac{\text{5}}{\tan \text{51,34}°}\\ y &= \text{4,00002...}\\ y &\approx \text{4} \end{aligned}$

Remember that the instructions say to round the answer to the second decimal place, as shown in the last step above.

The length of $\overline{ XV }$ is $\text{4}$ .

Submit your answer as:

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: $\; \tan47°$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

\tan 47 ° \approx

$\tan47° \approx$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate $\tan47°$ . If you are not sure about how to do that, refer to the manual of your calculator.

STEP: Use your calculator to evaluate

\tan 47 °

$\tan47°$ and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

\begin{aligned} \tan 47 ° & = 1,07236... \\ \approx 1,07 \end{aligned}

$\begin{aligned} \tan47° & = \text{1,07236...} \\ & \approx \text{1,07} \end{aligned}$

The correct answer is $\text{1,07}$ .

Submit your answer as:

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: $\; \tan75°$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

\tan 75 ° \approx

$\tan75° \approx$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate $\tan75°$ . If you are not sure about how to do that, refer to the manual of your calculator.

STEP: Use your calculator to evaluate

\tan 75 °

$\tan75°$ and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

\begin{aligned} \tan 75 ° & = 3,73205... \\ \approx 3,73 \end{aligned}

$\begin{aligned} \tan75° & = \text{3,73205...} \\ & \approx \text{3,73} \end{aligned}$

The correct answer is $\text{3,73}$ .

Submit your answer as:

Trigonometry: using a calculator

Use your calculator to evaluate the following expression: $\; \cos1°$ .

INSTRUCTION: Round your answer to two decimal places.

Answer:

\cos 1 ° \approx

$\cos1° \approx$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to use your calculator to evaluate $\cos1°$ . If you are not sure about how to do that, refer to the manual of your calculator.

STEP: Use your calculator to evaluate

\cos 1 °

$\cos1°$ and then round off to two decimal places
[−2 points ⇒ 0 / 2 points left]

To determine the answer, we must use a calculator.

\begin{aligned} \cos 1 ° & = 0,99984... \\ \approx 1 \end{aligned}

$\begin{aligned} \cos1° & = \text{0,99984...} \\ & \approx \text{1} \end{aligned}$

The correct answer is $\text{1}$ .

Submit your answer as:

Trigonometric ratios

Which trigonometric ratio is defined as the ratio of the opposite and the hypotenuse?

Answer: The trigonometric ratio is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Identify the correct trigonometric ratio
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The trigonometric ratios tell us about two sides and an angle in a right-angled triangle.

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

The diagram shows the two sides we need for the sine ratio in purple: opposite and hypotenuse. The adjacent is shaded out because it is not needed for the sine ratio.

Therefore, the ratio defined as the ratio of the opposite and the hypotenuse is sine.

Submit your answer as:

Trigonometric ratios

Which trigonometric ratio is defined as the ratio of the opposite and the hypotenuse?

Answer: The trigonometric ratio is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Identify the correct trigonometric ratio
[−1 point ⇒ 0 / 1 points left]

The trigonometric ratios tell us about two sides and an angle in a right-angled triangle.

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

The diagram shows the two sides we need for the sine ratio in purple: opposite and hypotenuse. The adjacent is shaded out because it is not needed for the sine ratio.

Therefore, the ratio defined as the ratio of the opposite and the hypotenuse is sine.

Submit your answer as:

Trigonometric ratios

Which trigonometric ratio is defined as the opposite over the hypotenuse?

Answer: The trigonometric ratio is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Identify the correct trigonometric ratio
[−1 point ⇒ 0 / 1 points left]

The trigonometric ratios tell us about two sides and an angle in a right-angled triangle.

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

The diagram shows the two sides we need for the sine ratio in purple: opposite and hypotenuse. The adjacent is shaded out because it is not needed for the sine ratio.

Therefore, the ratio defined as the opposite over the hypotenuse is sine.

Submit your answer as:

Trigonometric ratios: definitions

Complete the definition of sin using the dropdown menus below:

Answer:

Sin is defined as the over the .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Complete the definition of sin
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The trigonometric ratios tell us about two sides and an angle in a right-angled triangle.

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

The diagram shows the two sides we need for the sin ratio in purple: opposite and hypotenuse. The adjacent is shaded out because it is not needed for the sin ratio.

Therefore, sin is defined as the opposite over the hypotenuse.

Submit your answer as: and

Trigonometric ratios: definitions

Complete the definition of tan using the dropdown menus below:

Answer:

Tan is defined as the ratio of the and the .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Complete the definition of tan
[−2 points ⇒ 0 / 2 points left]

The trigonometric ratios tell us about two sides and an angle in a right-angled triangle.

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

The diagram shows the two sides we need for the tan ratio in purple: opposite and adjacent. The hypotenuse is shaded out because it is not needed for the tan ratio.

Therefore, tan is defined as the ratio of the opposite and the adjacent.

Submit your answer as: and

Trigonometric ratios: definitions

Complete the definition of sine using the dropdown menus below:

Answer:

Sine is defined as the ratio of the and the .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You can find the answer by reading about defining the trigonometric ratios in the Everything Maths Textbook.

STEP: Complete the definition of sine
[−2 points ⇒ 0 / 2 points left]

The trigonometric ratios tell us about two sides and an angle in a right-angled triangle.

We define the trigonometric ratios sine ( $\sin$ ), cosine ( $\cos$ ), and tangent ( $\tan$ ) of an angle $\theta$ as follows:

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \cos θ & = \frac{adjacent}{hypotenuse} \\ \tan θ & = \frac{opposite}{adjacent} \end{aligned}

The diagram shows the two sides we need for the sine ratio in purple: opposite and hypotenuse. The adjacent is shaded out because it is not needed for the sine ratio.

Therefore, sine is defined as the ratio of the opposite and the hypotenuse.

Submit your answer as: and

Writing trigonometric ratios

The diagram below shows a right-angled triangle, with sides $x$ , $y$ , and $z$ . The two acute angles are $A$ and $B$ . The diagram is not drawn to scale.

Determine the value of $\cos A$ for this triangle.

Answer:

\cos A =

$\cos A =$

expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides of the triangle. Remember that the opposite and adjacent depend on the position of the angle $A$ . Then use the definition of $\cos A$ to get the answer.

STEP: Evaluate

\cos A

$\cos A$ for the given triangle
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We have been given the lengths of the three sides of a right-angled triangle. We need to evaluate the ratio $\cos A$ . The definition of cosine is:

\begin{matrix} \cos A = \frac{adjacent}{hypotenuse} \end{matrix}

$\begin{array}{c}\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

We need to find the lengths for the adjacent and the hypotenuse. The adjacent side depends on the location of the angle $A$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the adjacent side. It is the side next to the angle $A$ .

In this case, the length of the adjacent is $x$ and the length of the hypotenuse is equal to $z$ . Therefore:

\begin{aligned} \cos A & = \frac{adjacent}{hypotenuse} \\ = \frac{x}{z} \end{aligned}

$\begin{aligned} \cos A & = \frac{\text{adjacent}}{\text{hypotenuse}} \\ & = \frac{x}{z} \end{aligned}$

Therefore, the answer is $\cos A = \dfrac{x}{z}$ .

Submit your answer as:

Writing trigonometric ratios

The diagram below shows a right-angled triangle, with sides $x$ , $y$ , and $z$ . The two acute angles are $\theta$ and $\lambda$ . The diagram is not drawn to scale.

Determine the value of $\cos \theta$ for this triangle.

Answer:

\cos θ =

$\cos \theta =$

expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides of the triangle. Remember that the opposite and adjacent depend on the position of the angle $\theta$ . Then use the definition of $\cos \theta$ to get the answer.

STEP: Evaluate

\cos θ

$\cos \theta$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to evaluate the ratio $\cos \theta$ . The definition of cosine is:

\begin{matrix} \cos θ = \frac{adjacent}{hypotenuse} \end{matrix}

$\begin{array}{c}\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

We need to find the lengths for the adjacent and the hypotenuse. The adjacent side depends on the location of the angle $\theta$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the adjacent side. It is the side next to the angle $\theta$ .

In this case, the length of the adjacent is $y$ and the length of the hypotenuse is equal to $z$ . Therefore:

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ = \frac{y}{z} \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \\ & = \frac{y}{z} \end{aligned}$

Therefore, the answer is $\cos \theta = \dfrac{y}{z}$ .

Submit your answer as:

Writing trigonometric ratios

The diagram below shows a right-angled triangle, with sides $p$ , $q$ , and $r$ . The two acute angles are $A$ and $B$ . The diagram is not drawn to scale.

Determine the value of $\sin A$ for this triangle.

Answer:

\sin A =

$\sin A =$

expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: Evaluate

\sin A

$\sin A$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to evaluate the ratio $\sin A$ . The definition of sine is:

\begin{matrix} \sin A = \frac{opposite}{hypotenuse} \end{matrix}

$\begin{array}{c}\sin A = \frac{\text{opposite}}{\text{hypotenuse}}\end{array}$

We need to find the lengths for the opposite and the hypotenuse. The opposite side depends on the location of the angle $A$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the opposite side. It is the side across the triangle from the angle $A$ .

In this case, the length of the opposite is $p$ and the length of the hypotenuse is equal to $q$ . Therefore:

\begin{aligned} \sin A & = \frac{opposite}{hypotenuse} \\ = \frac{p}{q} \end{aligned}

$\begin{aligned} \sin A & = \frac{\text{opposite}}{\text{hypotenuse}} \\ & = \frac{p}{q} \end{aligned}$

Therefore, the answer is $\sin A = \dfrac{p}{q}$ .

Submit your answer as:

Trigonometric ratios: opposite and adjacent sides

Consider the diagram below:

The following equation is based on the triangle above.

\begin{aligned} \sin 37 ° = \frac{S}{◻} \end{aligned}

$\begin{aligned} \sin 37° = \frac{S}{\Box} \end{aligned}$

What is the quantity represented by $\Box$ ?

Answer:

The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by identifying the hypotenuse, adjacent, and opposite sides of the right-angled triangle given. Then use the following definition to answer the question:

\begin{matrix} \sin 37 ° = \frac{opposite}{hypotenuse} \end{matrix}

$\begin{array}{c}\sin 37° = \frac{\text{opposite}}{\text{hypotenuse}}\end{array}$

STEP: Use the definition of

\sin

$\sin$

37 °

$\text{37}\text{°}$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

\begin{aligned} \sin 37 ° = \frac{S}{◻} \end{aligned}

$\begin{aligned} \sin 37° = \frac{S}{\Box} \end{aligned}$

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

side $Q$ is the adjacent side because it is next to (or adjacent to) the $\text{37}\text{°}$ angle.
side $S$ is the opposite side because it is across from (or opposite to) the $\text{37}\text{°}$ angle.

In order to define the sine of an acute angle in a right-angled triangle, we make use of the opposite and hypotenuse sides. In this case:

\begin{matrix} \sin 37 ° = \frac{opposite}{hypotenuse} \end{matrix}

$\begin{array}{c}\sin 37° = \frac{\text{opposite}}{\text{hypotenuse}}\end{array}$

In the triangle diagram given, the label of the opposite side is $S$ and the label for the hypotenuse side is $R$ . We have drawn circles around these labels in the diagram shown below:

NOTE: We have shaded the side

Q

$Q$ because it is not required in the definition of the sine of the angle

37 °

$\text{37}\text{°}$ .

We can now use these labels to define sine for the angle $\text{37}\text{°}$ :

\begin{aligned} \sin 37 ° = \frac{S}{R} \end{aligned}

$\begin{aligned}\sin 37° = \frac{S}{R}\end{aligned}$

Comparing this to the equation with the unknown quantity, we can see that the identity of $\Box$ is the quantity $R$ .

The missing quantity $\Box$ is $R$ .

Submit your answer as:

Trigonometric ratios: opposite and adjacent sides

Consider the diagram below:

The following equation is based on the triangle above.

\begin{aligned} \cos 39 ° = \frac{S}{◻} \end{aligned}

$\begin{aligned} \cos 39° = \frac{S}{\Box} \end{aligned}$

What is the quantity represented by $\Box$ ?

Answer:

The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by identifying the hypotenuse, adjacent, and opposite sides of the right-angled triangle given. Then use the following definition to answer the question:

\begin{matrix} \cos 39 ° = \frac{adjacent}{hypotenuse} \end{matrix}

$\begin{array}{c}\cos 39° = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

STEP: Use the definition of

\cos

$\cos$

39 °

$\text{39}\text{°}$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

\begin{aligned} \cos 39 ° = \frac{S}{◻} \end{aligned}

$\begin{aligned} \cos 39° = \frac{S}{\Box} \end{aligned}$

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

side $S$ is the adjacent side because it is next to (or adjacent to) the $\text{39}\text{°}$ angle.
side $Q$ is the opposite side because it is across from (or opposite to) the $\text{39}\text{°}$ angle.

In order to define the cosine of an acute angle in a right-angled triangle, we make use of the adjacent and hypotenuse sides. In this case:

\begin{matrix} \cos 39 ° = \frac{adjacent}{hypotenuse} \end{matrix}

$\begin{array}{c}\cos 39° = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

In the triangle diagram given, the label of the adjacent side is $S$ and the label for the hypotenuse side is $R$ . We have drawn circles around these labels in the diagram shown below:

NOTE: We have shaded the side

Q

$Q$ because it is not required in the definition of the cosine of the angle

39 °

$\text{39}\text{°}$ .

We can now use these labels to define cosine for the angle $\text{39}\text{°}$ :

\begin{aligned} \cos 39 ° = \frac{S}{R} \end{aligned}

$\begin{aligned}\cos 39° = \frac{S}{R}\end{aligned}$

Comparing this to the equation with the unknown quantity, we can see that the identity of $\Box$ is the quantity $R$ .

The missing quantity $\Box$ is $R$ .

Submit your answer as:

Trigonometric ratios: opposite and adjacent sides

Consider the diagram below:

The following equation is based on the triangle above.

\begin{aligned} \sin 69 ° = \frac{◻}{R} \end{aligned}

$\begin{aligned} \sin 69° = \frac{\Box}{R} \end{aligned}$

What is the quantity represented by $\Box$ ?

Answer:

The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by identifying the hypotenuse, adjacent, and opposite sides of the right-angled triangle given. Then use the following definition to answer the question:

\begin{matrix} \sin 69 ° = \frac{opposite}{hypotenuse} \end{matrix}

$\begin{array}{c}\sin 69° = \frac{\text{opposite}}{\text{hypotenuse}}\end{array}$

STEP: Use the definition of

\sin

$\sin$

69 °

$\text{69}\text{°}$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

\begin{aligned} \sin 69 ° = \frac{◻}{R} \end{aligned}

$\begin{aligned} \sin 69° = \frac{\Box}{R} \end{aligned}$

The labels of the sides of a right-angled triangle are: adjacent, opposite, and hypotenuse.

side $S$ is the adjacent side because it is next to (or adjacent to) the $\text{69}\text{°}$ angle.
side $Q$ is the opposite side because it is across from (or opposite to) the $\text{69}\text{°}$ angle.

In order to define the sine of an acute angle in a right-angled triangle, we make use of the opposite and hypotenuse sides. In this case:

\begin{matrix} \sin 69 ° = \frac{opposite}{hypotenuse} \end{matrix}

$\begin{array}{c}\sin 69° = \frac{\text{opposite}}{\text{hypotenuse}}\end{array}$

In the triangle diagram given, the label of the opposite side is $Q$ and the label for the hypotenuse side is $R$ . We have drawn circles around these labels in the diagram shown below:

NOTE: We have shaded the side

S

$S$ because it is not required in the definition of the sine of the angle

69 °

$\text{69}\text{°}$ .

We can now use these labels to define sine for the angle $\text{69}\text{°}$ :

\begin{aligned} \sin 69 ° = \frac{Q}{R} \end{aligned}

$\begin{aligned}\sin 69° = \frac{Q}{R}\end{aligned}$

Comparing this to the equation with the unknown quantity, we can see that the identity of $\Box$ is the quantity $Q$ .

The missing quantity $\Box$ is $Q$ .

Submit your answer as:

Trigonometric ratios

The following diagram shows a right-angled triangle. The sides are labelled with lengths $10$ , $5 \sqrt{5}$ , and $15$ . The two acute angles are labelled $\alpha$ and $\beta$ . The diagram is not drawn to scale.

Determine the value of $\sin \alpha$ in the following triangle.

INSTRUCTION: Give your answer in surd form, if necessary, and simplify your answer completely. Type sqrt( ) if you need to indicate a square root.

Answer:

\sin α =

$\sin \alpha =$

expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides of the triangle. Then use the definition of $\sin \alpha$ to get the answer.

STEP: Determine the value of

\sin α

$\sin \alpha$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to determine the value of $\sin \alpha$ .

To find $\sin \alpha$ , we need to identify the opposite and the hypotenuse. The opposite side depends on the location of the angle $\alpha$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the the opposite side. It is the side across the triangle form the angle $\alpha$ .

In this case, the length of the opposite is $5 \sqrt{5}$ and the length of the hypotenuse is equal to $15$ . Therefore:

\begin{aligned} \sin α & = \frac{5 \sqrt{5}}{15} \\ = \frac{\sqrt{5}}{3} \end{aligned}

$\begin{aligned} \sin \alpha & = \frac{5 \sqrt{5}}{15} \\ & = \frac{\sqrt{5}}{3} \end{aligned}$

Therefore, $\sin \alpha = \dfrac{\sqrt{5}}{3}$ .

Submit your answer as:

Trigonometric ratios

The following diagram shows a right-angled triangle. The sides are labelled with lengths $2 \sqrt{13}$ , $12$ , and $14$ . The two acute angles are labelled $\theta$ and $\lambda$ . The diagram is not drawn to scale.

Determine the value of $\cos \theta$ in the following triangle.

INSTRUCTION: Give your answer in surd form, if necessary, and simplify your answer completely. Type sqrt( ) if you need to indicate a square root.

Answer:

\cos θ =

$\cos \theta =$

expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides of the triangle. Then use the definition of $\cos \theta$ to get the answer.

STEP: Determine the value of

\cos θ

$\cos \theta$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to determine the value of $\cos \theta$ .

To find $\cos \theta$ , we need to identify the adjacent and the hypotenuse. The adjacent side depends on the location of the angle $\theta$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the the adjacent side. It is the side next to the angle $\theta$ .

In this case, the length of the adjacent is $2 \sqrt{13}$ and the length of the hypotenuse is equal to $14$ . Therefore:

\begin{aligned} \cos θ & = \frac{2 \sqrt{13}}{14} \\ = \frac{\sqrt{13}}{7} \end{aligned}

$\begin{aligned} \cos \theta & = \frac{2 \sqrt{13}}{14} \\ & = \frac{\sqrt{13}}{7} \end{aligned}$

Therefore, $\cos \theta = \dfrac{\sqrt{13}}{7}$ .

Submit your answer as:

Trigonometric ratios

The following diagram shows a right-angled triangle. The sides are labelled with lengths $10 \sqrt{3}$ , $10$ , and $20$ . The two acute angles are labelled $A$ and $B$ . The diagram is not drawn to scale.

Determine the value of $\tan B$ in the following triangle.

INSTRUCTION: Give your answer in surd form, if necessary, and simplify your answer completely. Type sqrt( ) if you need to indicate a square root.

Answer:

\tan B =

$\tan B =$

expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides of the triangle. Then use the definition of $\tan B$ to get the answer.

STEP: Determine the value of

\tan B

$\tan B$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to determine the value of $\tan B$ .

To find $\tan B$ , we need to identify the opposite and the adjacent. The opposite side is across the triangle from the angle $B$ . The adjacent side is next to $B$ . The arrows and circles below show the those two sides. The opposite side is across the triangle from $B$ , and the adjacent side is next to $B$ .

In this case, the length of the opposite is $10$ and the length of the adjacent is equal to $10 \sqrt{3}$ . Therefore:

\begin{aligned} \tan B & = \frac{10}{10 \sqrt{3}} \\ = \frac{2}{2 \sqrt{3}} \\ = \frac{\sqrt{3}}{3} \end{aligned}

$\begin{aligned} \tan B & = \frac{10}{10 \sqrt{3}} \\ & = \frac{2}{2 \sqrt{3}} \\ & = \frac{\sqrt{3}}{3} \end{aligned}$

Therefore, $\tan B = \dfrac{\sqrt{3}}{3}$ .

Submit your answer as:

Trigonometric ratio definitions

The triangle $MNP$ below is right-angled at $M$ .

Use the triangle $MNP$ to answer the following questions:

What is the quantity represented by $\Box$ in the following equation?

$\begin{aligned} \tan \hat{N} = \frac{M P}{◻} \end{aligned}$ $\begin{aligned} \tan \hat{N} = \frac{MP}{\Box} \end{aligned}$
Answer:
The missing quantity is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the angle $\hat{N}$ to identify the hypotenuse, adjacent, and opposite sides of the triangle $MNP$ . After that, use the following definition to answer the question:

$\begin{matrix} \tan \hat{N} = \frac{opposite}{adjacent} \end{matrix}$ $\begin{array}{c}\tan \hat{N} = \frac{\text{opposite}}{\text{adjacent}}\end{array}$

STEP: Use the definition of $\tan\hat{N}$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

$\begin{aligned} \tan \hat{N} = \frac{M P}{◻} \end{aligned}$ $\begin{aligned} \tan \hat{N} = \frac{MP}{\Box} \end{aligned}$

The names of the sides of a right-angled triangle are hypotenuse, adjacent, and opposite sides. The side that does not touch the angle $\hat{M}$ is the hypotenuse. In this question, the hypotenuse is the side $NP$ . The names of the other two sides depend on the angle of interest. In this question, the angle of interest is angle $\hat{N}$ . Using this angle:
- side $MN$ is the adjacent side because it is next to (or adjacent to) the angle $\hat{N}$ .
- side $MP$ is the opposite side because it is across from (or opposite to) the angle $\hat{N}$ .
To define tangent we make use of the sides: opposite and adjacent. The definition of tangent for the angle $\hat{N}$ is:

$\begin{matrix} \tan \hat{N} = \frac{opposite}{adjacent} \end{matrix}$ $\begin{array}{c}\tan \hat{N} = \frac{\text{opposite}}{\text{adjacent}}\end{array}$

From the triangle diagram, the label for the opposite side is $MP$ , and the label for the adjacent side is $MN$ .

Comparing this to the equation with the unknown quantity, we can see that the identity of $\Box$ is the quantity $MN$ . We can now complete the equation:

$\begin{aligned} \tan \hat{N} = \frac{M P}{M N} \end{aligned}$ $\begin{aligned} \tan \hat{N} = \frac{MP}{MN} \end{aligned}$

NOTE:
We will redraw the triangle $MNP$ to show the two sides we need for the tangent of the angle $\hat{N}$ . These are: $MP$ and $MN$ .

The hypotenuse side, $NP$ , has been shaded out because it is not needed for the tangent ratio.

The missing quantity is $MN$ .

Submit your answer as:
From the drop-down list below, choose the quantity that would make the following equation complete.

$\begin{aligned} ◻ \hat{P} = \frac{M N}{N P} \end{aligned}$ $\begin{aligned} \Box \; \hat{P} = \frac{MN}{NP} \end{aligned}$

Answer: The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The missing quantity is a trigonometric function. Using the angle $\hat{P}$ , identify the hypotenuse, adjacent, and opposite sides of the triangle $MNP$ . Then use any of the following definitions to answer the question:

$\begin{aligned} \cos \hat{P} & = \frac{adjacent}{hypotenuse} \\ \sin \hat{P} & = \frac{opposite}{hypotenuse} \\ \tan \hat{P} & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{array}{rl} \cos \hat{P} &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \hat{P} &= \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \tan \hat{P} &= \dfrac{\text{opposite}}{\text{adjacent}} \end{array}$

STEP: Identify the trigonometric ratio of the angle $\hat{P}$ equal to the ratio $\dfrac{MN}{NP}$
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

$\begin{aligned} ◻ \hat{P} = \frac{M N}{N P} \end{aligned}$ $\begin{aligned} \Box \; \hat{P} = \frac{MN}{NP} \end{aligned}$

The missing quantity is a trigonometric function. Trigonometric functions of acute angles of a right-angled triangle are equal to the ratio of two of the sides. The definitions of three trigonometric functions of the angle $\hat{P}$ using the names of the sides, are:

$\begin{aligned} \cos \hat{P} & = \frac{adjacent}{hypotenuse} \\ \sin \hat{P} & = \frac{opposite}{hypotenuse} \\ \tan \hat{P} & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{array}{rl} \cos \hat{P} &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \hat{P} &= \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \tan \hat{P} &= \dfrac{\text{opposite}}{\text{adjacent}} \end{array}$

We now need to identify the names of the sides: $MN$ and $NP$ . Then we will identify the missing trigonometric function.

For the angle $\hat{P}$ , the adjacent is the side $MP$ , the opposite is side $MN$ , and the hypotenuse is side $NP$ . The equation in the question statement has the ratio of the opposite side to the hypotenuse side. Using the equations of the definitions of the three trigonometric ratios above, we find that the missing quantity is $\sin$ . The completed equation is:

$\begin{aligned} \sin \hat{P} = \frac{M N}{N P} \end{aligned}$ $\begin{aligned} \sin \hat{P} = \frac{MN}{NP} \end{aligned}$

NOTE:
We will redraw the triangle $MNP$ to show the two sides we need for the sine of the angle $\hat{P}$ . These are: the opposite side $MN$ and the hypotenuse side $NP$ .

The missing quantity is $\sin$ .

Submit your answer as:

Trigonometric ratio definitions

The triangle $PQR$ below is right-angled at $P$ .

Use the triangle $PQR$ to answer the following questions:

What is the quantity represented by $\Box$ in the following equation?

$\begin{aligned} \sin \hat{R} = \frac{P Q}{◻} \end{aligned}$ $\begin{aligned} \sin \hat{R} = \frac{PQ}{\Box} \end{aligned}$
Answer:
The missing quantity is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the angle $\hat{R}$ to identify the hypotenuse, adjacent, and opposite sides of the triangle $PQR$ . After that, use the following definition to answer the question:

$\begin{matrix} \sin \hat{R} = \frac{opposite}{hypotenuse} \end{matrix}$ $\begin{array}{c}\sin \hat{R} = \frac{\text{opposite}}{\text{hypotenuse}}\end{array}$

STEP: Use the definition of $\sin\hat{R}$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

$\begin{aligned} \sin \hat{R} = \frac{P Q}{◻} \end{aligned}$ $\begin{aligned} \sin \hat{R} = \frac{PQ}{\Box} \end{aligned}$

The names of the sides of a right-angled triangle are hypotenuse, adjacent, and opposite sides. The side that does not touch the angle $\hat{P}$ is the hypotenuse. In this question, the hypotenuse is the side $QR$ . The names of the other two sides depend on the angle of interest. In this question, the angle of interest is angle $\hat{R}$ . Using this angle:
- side $PR$ is the adjacent side because it is next to (or adjacent to) the angle $\hat{R}$ .
- side $PQ$ is the opposite side because it is across from (or opposite to) the angle $\hat{R}$ .
To define sine we make use of the sides: opposite and hypotenuse. The definition of sine for the angle $\hat{R}$ is:

$\begin{matrix} \sin \hat{R} = \frac{opposite}{hypotenuse} \end{matrix}$ $\begin{array}{c}\sin \hat{R} = \frac{\text{opposite}}{\text{hypotenuse}}\end{array}$

From the triangle diagram, the label for the opposite side is $PQ$ , and the label for the hypotenuse side is $QR$ .

Comparing this to the equation with the unknown quantity, we can see that the identity of $\Box$ is the quantity $QR$ . We can now complete the equation:

$\begin{aligned} \sin \hat{R} = \frac{P Q}{Q R} \end{aligned}$ $\begin{aligned} \sin \hat{R} = \frac{PQ}{QR} \end{aligned}$

NOTE:
We will redraw the triangle $PQR$ to show the two sides we need for the sine of the angle $\hat{R}$ . These are: $PQ$ and $QR$ .

The adjacent side, $PR$ , has been shaded out because it is not needed for the sine ratio.

The missing quantity is $QR$ .

Submit your answer as:
From the drop-down list below, choose the quantity that would make the following equation complete.

$\begin{aligned} ◻ \hat{R} = \frac{P Q}{P R} \end{aligned}$ $\begin{aligned} \Box \; \hat{R} = \frac{PQ}{PR} \end{aligned}$

Answer: The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The missing quantity is a trigonometric function. Using the angle $\hat{R}$ , identify the hypotenuse, adjacent, and opposite sides of the triangle $PQR$ . Then use any of the following definitions to answer the question:

$\begin{aligned} \cos \hat{R} & = \frac{adjacent}{hypotenuse} \\ \sin \hat{R} & = \frac{opposite}{hypotenuse} \\ \tan \hat{R} & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{array}{rl} \cos \hat{R} &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \hat{R} &= \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \tan \hat{R} &= \dfrac{\text{opposite}}{\text{adjacent}} \end{array}$

STEP: Identify the trigonometric ratio of the angle $\hat{R}$ equal to the ratio $\dfrac{PQ}{PR}$
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

$\begin{aligned} ◻ \hat{R} = \frac{P Q}{P R} \end{aligned}$ $\begin{aligned} \Box \; \hat{R} = \frac{PQ}{PR} \end{aligned}$

The missing quantity is a trigonometric function. Trigonometric functions of acute angles of a right-angled triangle are equal to the ratio of two of the sides. The definitions of three trigonometric functions of the angle $\hat{R}$ using the names of the sides, are:

$\begin{aligned} \cos \hat{R} & = \frac{adjacent}{hypotenuse} \\ \sin \hat{R} & = \frac{opposite}{hypotenuse} \\ \tan \hat{R} & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{array}{rl} \cos \hat{R} &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \hat{R} &= \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \tan \hat{R} &= \dfrac{\text{opposite}}{\text{adjacent}} \end{array}$

We now need to identify the names of the sides: $PQ$ and $PR$ . Then we will identify the missing trigonometric function.

For the angle $\hat{R}$ , the adjacent is the side $PR$ , the opposite is side $PQ$ , and the hypotenuse is side $QR$ . The equation in the question statement has the ratio of the opposite side to the adjacent side. Using the equations of the definitions of the three trigonometric ratios above, we find that the missing quantity is $\tan$ . The completed equation is:

$\begin{aligned} \tan \hat{R} = \frac{P Q}{P R} \end{aligned}$ $\begin{aligned} \tan \hat{R} = \frac{PQ}{PR} \end{aligned}$

NOTE:
We will redraw the triangle $PQR$ to show the two sides we need for the tangent of the angle $\hat{R}$ . These are: the opposite side $PQ$ and the adjacent side $PR$ .

The missing quantity is $\tan$ .

Submit your answer as:

Trigonometric ratio definitions

The triangle $MNP$ below is right-angled at $M$ .

Use the triangle $MNP$ to answer the following questions:

What is the quantity represented by $\Box$ in the following equation?

$\begin{aligned} \cos \hat{N} = \frac{M N}{◻} \end{aligned}$ $\begin{aligned} \cos \hat{N} = \frac{MN}{\Box} \end{aligned}$
Answer:
The missing quantity is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the angle $\hat{N}$ to identify the hypotenuse, adjacent, and opposite sides of the triangle $MNP$ . After that, use the following definition to answer the question:

$\begin{matrix} \cos \hat{N} = \frac{adjacent}{hypotenuse} \end{matrix}$ $\begin{array}{c}\cos \hat{N} = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

STEP: Use the definition of $\cos\hat{N}$ to get the answer
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

$\begin{aligned} \cos \hat{N} = \frac{M N}{◻} \end{aligned}$ $\begin{aligned} \cos \hat{N} = \frac{MN}{\Box} \end{aligned}$

The names of the sides of a right-angled triangle are hypotenuse, adjacent, and opposite sides. The side that does not touch the angle $\hat{M}$ is the hypotenuse. In this question, the hypotenuse is the side $NP$ . The names of the other two sides depend on the angle of interest. In this question, the angle of interest is angle $\hat{N}$ . Using this angle:
- side $MN$ is the adjacent side because it is next to (or adjacent to) the angle $\hat{N}$ .
- side $MP$ is the opposite side because it is across from (or opposite to) the angle $\hat{N}$ .
To define cosine we make use of the sides: adjacent and hypotenuse. The definition of cosine for the angle $\hat{N}$ is:

$\begin{matrix} \cos \hat{N} = \frac{adjacent}{hypotenuse} \end{matrix}$ $\begin{array}{c}\cos \hat{N} = \frac{\text{adjacent}}{\text{hypotenuse}}\end{array}$

From the triangle diagram, the label for the adjacent side is $MN$ , and the label for the hypotenuse side is $NP$ .

Comparing this to the equation with the unknown quantity, we can see that the identity of $\Box$ is the quantity $NP$ . We can now complete the equation:

$\begin{aligned} \cos \hat{N} = \frac{M N}{N P} \end{aligned}$ $\begin{aligned} \cos \hat{N} = \frac{MN}{NP} \end{aligned}$

NOTE:
We will redraw the triangle $MNP$ to show the two sides we need for the cosine of the angle $\hat{N}$ . These are: $MN$ and $NP$ .

The opposite side, $MP$ , has been shaded out because it is not needed for the cosine ratio.

The missing quantity is $NP$ .

Submit your answer as:
From the drop-down list below, choose the quantity that would make the following equation complete.

$\begin{aligned} ◻ \hat{N} = \frac{M P}{N P} \end{aligned}$ $\begin{aligned} \Box \; \hat{N} = \frac{MP}{NP} \end{aligned}$

Answer: The missing quantity is .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The missing quantity is a trigonometric function. Using the angle $\hat{N}$ , identify the hypotenuse, adjacent, and opposite sides of the triangle $MNP$ . Then use any of the following definitions to answer the question:

$\begin{aligned} \cos \hat{N} & = \frac{adjacent}{hypotenuse} \\ \sin \hat{N} & = \frac{opposite}{hypotenuse} \\ \tan \hat{N} & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{array}{rl} \cos \hat{N} &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \hat{N} &= \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \tan \hat{N} &= \dfrac{\text{opposite}}{\text{adjacent}} \end{array}$

STEP: Identify the trigonometric ratio of the angle $\hat{N}$ equal to the ratio $\dfrac{MP}{NP}$
[−1 point ⇒ 0 / 1 points left]

We need to find the missing quantity in the equation:

$\begin{aligned} ◻ \hat{N} = \frac{M P}{N P} \end{aligned}$ $\begin{aligned} \Box \; \hat{N} = \frac{MP}{NP} \end{aligned}$

The missing quantity is a trigonometric function. Trigonometric functions of acute angles of a right-angled triangle are equal to the ratio of two of the sides. The definitions of three trigonometric functions of the angle $\hat{N}$ using the names of the sides, are:

$\begin{aligned} \cos \hat{N} & = \frac{adjacent}{hypotenuse} \\ \sin \hat{N} & = \frac{opposite}{hypotenuse} \\ \tan \hat{N} & = \frac{opposite}{adjacent} \end{aligned}$ $\begin{array}{rl} \cos \hat{N} &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\ \sin \hat{N} &= \dfrac{\text{opposite}}{\text{hypotenuse}} \\ \tan \hat{N} &= \dfrac{\text{opposite}}{\text{adjacent}} \end{array}$

We now need to identify the names of the sides: $MP$ and $NP$ . Then we will identify the missing trigonometric function.

For the angle $\hat{N}$ , the adjacent is the side $MN$ , the opposite is side $MP$ , and the hypotenuse is side $NP$ . The equation in the question statement has the ratio of the opposite side to the hypotenuse side. Using the equations of the definitions of the three trigonometric ratios above, we find that the missing quantity is $\sin$ . The completed equation is:

$\begin{aligned} \sin \hat{N} = \frac{M P}{N P} \end{aligned}$ $\begin{aligned} \sin \hat{N} = \frac{MP}{NP} \end{aligned}$

NOTE:
We will redraw the triangle $MNP$ to show the two sides we need for the sine of the angle $\hat{N}$ . These are: the opposite side $MP$ and the hypotenuse side $NP$ .

The missing quantity is $\sin$ .

Submit your answer as:

Evaluating trigonometric ratios

The diagram below shows a right-angled triangle with sides of length $40$ , $30$ , and $50$ . The two acute angles are labelled $\theta$ and $\lambda$ .

Determine the value of $\tan \theta$ for this triangle.

INSTRUCTION: Give your answer as a simplified fraction.

Answer:

\tan θ =

$\tan \theta =$

fraction

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the definition of the $\tan$ ratio. Then identify which sides of the triangle correspond to the sides in the definition to find the answer.

STEP: Determine the value of

\tan θ

$\tan \theta$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to evaluate the ratio $\tan \theta$ .

The definition of $\tan \theta$ is

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \end{aligned}

$\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} \end{aligned}$

So we need to identify the opposite and adjacent sides for the angle $\theta$ . The opposite side is across the triangle from the angle $\theta$ . The adjacent side is next to $\theta$ . The arrows and circles below show those two sides. The opposite side is across the triangle from $\theta$ , and the adjacent side is next to $\theta$ .

In this case, the length of the opposite is $\text{30}$ and the length of the adjacent is $\text{40}$ . Therefore:

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ = \frac{30}{40} \\ = \frac{3}{4} \end{aligned}

$\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} \\ & = \frac{30}{40} \\ & = \frac{3}{4} \end{aligned}$

Therefore, the answer is $\tan \theta = \dfrac{3}{4}$ .

Submit your answer as:

Evaluating trigonometric ratios

The diagram below shows a right-angled triangle with sides of length $16$ , $34$ , and $30$ . The two acute angles are labelled $\alpha$ and $\beta$ .

Determine the value of $\sin \alpha$ for this triangle.

INSTRUCTION: Give your answer as a simplified fraction.

Answer:

\sin α =

$\sin \alpha =$

fraction

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the definition of the $\sin$ ratio. Then identify which sides of the triangle correspond to the sides in the definition to find the answer.

STEP: Determine the value of

\sin α

$\sin \alpha$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to evaluate the ratio $\sin \alpha$ .

The definition of $\sin \alpha$ is

\begin{aligned} \sin α & = \frac{opposite}{hypotenuse} \end{aligned}

$\begin{aligned} \sin \alpha & = \frac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$

So we need to identify the opposite and hypotenuse sides for the angle $\alpha$ . The opposite side depends on the location of the angle $\alpha$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the opposite side. It is the side across the triangle from the angle $\alpha$ .

In this case, the length of the opposite is $\text{30}$ and the length of the hypotenuse is $\text{34}$ . Therefore:

\begin{aligned} \sin α & = \frac{opposite}{hypotenuse} \\ = \frac{30}{34} \\ = \frac{15}{17} \end{aligned}

$\begin{aligned} \sin \alpha & = \frac{\text{opposite}}{\text{hypotenuse}} \\ & = \frac{30}{34} \\ & = \frac{15}{17} \end{aligned}$

Therefore, the answer is $\sin \alpha = \dfrac{15}{17}$ .

Submit your answer as:

Evaluating trigonometric ratios

The diagram below shows a right-angled triangle with sides of length $24$ , $18$ , and $30$ . The two acute angles are labelled $A$ and $B$ .

Determine the value of $\sin A$ for this triangle.

INSTRUCTION: Give your answer as a simplified fraction.

Answer:

\sin A =

$\sin A =$

fraction

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the definition of the $\sin$ ratio. Then identify which sides of the triangle correspond to the sides in the definition to find the answer.

STEP: Determine the value of

\sin A

$\sin A$ for the given triangle
[−2 points ⇒ 0 / 2 points left]

We have been given the lengths of the three sides of a right-angled triangle. We need to evaluate the ratio $\sin A$ .

The definition of $\sin A$ is

\begin{aligned} \sin A & = \frac{opposite}{hypotenuse} \end{aligned}

$\begin{aligned} \sin A & = \frac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$

So we need to identify the opposite and hypotenuse sides for the angle $A$ . The opposite side depends on the location of the angle $A$ . (The hypotenuse is always the side across from the right-angle.) The arrow and circle below show the opposite side. It is the side across the triangle from the angle $A$ .

In this case, the length of the opposite is $\text{18}$ and the length of the hypotenuse is $\text{30}$ . Therefore:

\begin{aligned} \sin A & = \frac{opposite}{hypotenuse} \\ = \frac{18}{30} \\ = \frac{3}{5} \end{aligned}

$\begin{aligned} \sin A & = \frac{\text{opposite}}{\text{hypotenuse}} \\ & = \frac{18}{30} \\ & = \frac{3}{5} \end{aligned}$

Therefore, the answer is $\sin A = \dfrac{3}{5}$ .

Submit your answer as:

2. Practical applications

Solving two dimensional problems

A person stands at Point A, looking up at a bird sitting on the top of a building, Point B.

The height of the building is $7$ meters, Point A is $x$ meters away from the foot of the building, and the angle of incline (from the horizontal) the the top of the building is $65°$ .

Calculate the distance away from the foot of the building ( $x$ ) as shown in the diagram below. Round your answer to two decimal places where necessary.

Answer:

x

$x$ =

m

$\text{m}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the tangent trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the correct trig ratio to use is the tangent function.

\begin{aligned} \tan (θ) & = \frac{opposite}{adjacent} \\ \tan (65) ° & = \frac{7}{x} \\ x & = \frac{7}{\tan (65) °} \\ x & = 3,2641... \\ x & \approx 3, 26 \end{aligned}

$\begin{aligned} \tan (\theta) &= \frac{\text{opposite}}{\text{adjacent}}\\ \tan (65)° &= \frac{\text{7}}{x}\\ x &= \frac{\text{7}}{\tan(65)°} \\ x &= \text{3,2641...}\\ x &\approx 3,26 \end{aligned}$

Remember to round your answer to two decimal places. Therefore $x$ is $\text{3,26}\ \text{m}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point $A$ , looking up at a bird sitting on the top of a building, at Point $B$ .

The height of the building is $\text{5}\ \text{m}$ , the line-of-sight distance from Point $A$ to the top of the building (Point $B$ ) is $x$ , and the angle of inclination (from the horizontal) to the top of the building is $\text{61}°$ .

Calculate the line-of-sight distance to the top of the building ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer: The line-of-sight distance to the top of the building is

m

$\text{m}$ .

one-of

type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the sine trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the trig ratio to use is $\sin$ .

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin 61 ° & = \frac{5}{x} \\ x & = \frac{5}{\sin 61 °} \\ x & = 5,71677... \\ \approx 5,72 m \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin \text{61}\text{°} & = \frac{\text{5}}{x} \\ x & = \frac{\text{5}}{\sin\text{61}\text{°}}\\ x & = \text{5,71677...}\\ & \approx \text{5,72}\ \text{m} \end{aligned}$

The line-of-sight distance to the top of the building is $\text{5,72}\ \text{m}$ .

Submit your answer as:

Using trigonometry with word problems

A man stands at a window on the first floor of a building, shouting down to a child on the other side of the street. The following are true:

The man in the window is $\text{3}\ \text{m}$ above the street.
The horizontal distance from the man standing in the window to the point directly above the child on the street is $\text{4,5}\ \text{m}$ .

Find the angle of depression from the man down to the child.

INSTRUCTION: Round your answer to two decimal places.

Answer: The angle is °.

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by drawing a triangle diagram to represent the situation in the question. Label all of the information given in the question. Also label any other information you know is true (for example, if the triangle is equilateral, mark all three sides to show they are the same length).

STEP: Draw a diagram to represent the situation and label it
[−1 point ⇒ 3 / 4 points left]

When you have a word problem about a shape or positions and there is no diagram, start by drawing a sketch to summarise the question. Label the sketch with all of the information given. In this case, the information given in the question fits on a triangle: there is one thing at ground level (the child on the street) and something else up above (the man in the window). We can represent this with a right-angled triangle.

The picture does not need to be perfect but try to make it reasonable. For example, the question gives you two lengths, one of which is longer than the other: draw the picture so that the longer side is clearly longer than the shorter length.

NOTE: The diagram should also have a label for what you want to find. We have used "

a

$a$ " to label the angle that we want to find (but you can use any variable that you want).

STEP: Determine which trigonometric ratio to use
[−1 point ⇒ 2 / 4 points left]

One of the most important facts about the picture above is that it is a right-angled triangle because the street and the building are perpendicular to each other. That means we can use the trigonometric ratios to try to solve this question, or even the theorem of Pythagoras: we should pick whichever option will get us the answer.

In this case, the sides given are the two sides opposite from and adjacent to the angle we want. Therefore, the correct trigonometric ratio is the tangent function. (In fact, there is no other way to solve the question with the information given. While the theorem of Pythagoras applies to the triangle, it is about the lengths of the sides, and we need to find an angle: the theorem is useless for this question.)

STEP: Set up an equation and solve for the answer
[−2 points ⇒ 0 / 4 points left]

Now we can write an equation for the tangent ratio from the sketch, and solve for the angle we need. Start by writing the equation for tangent, and then substitute in the correct values.

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ \tan a & = \frac{3}{4,5} \\ a & = \tan^{- 1} (\frac{3}{4,5}) \\ a & = 33,6900... \\ a & \approx 33,69 ° \end{aligned}

$\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} \\ \tan a & = \frac{\text{3}}{\text{4,5}}\\ a & = \tan^{-1} \left( \frac{\text{3}}{\text{4,5}} \right) \\ a & = \text{33,6900...} \\ a & \approx \text{33,69}\text{°} \end{aligned}$

We got it! As shown in the last step, we rounded off to the nearest hundredths place because the question told us to.

The angle of depression from the man down to the child is $\text{33,69}\text{°}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point $A$ , looking up at a bird sitting on the top of a tree, Point $B$ .

Point $A$ is $9.0$ meters away from the foot of the tree, the line-of-sight distance from Point $A$ to the top of the tree (Point $B$ ) is $x$ meters, and the angle of incline (from the horizontal) the the top of the tree is $\text{45}\text{°}$ .

Calculate the line-of-sight distance to the top of the tree ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x=$

m

$\text{m}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the cosine trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the trig ratio to use is $\cos$ .

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos 45 ° & = \frac{9.0}{x} \\ ∴ x & = 12,73 m \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \text{45}° & = \frac{9.0}{x} \\ \therefore x & = \text{12,73}\ \text{m} \end{aligned}$

Therefore $x$ is $\text{12,73}\ \text{m}$ .

Submit your answer as:

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows an isosceles triangle. Side $\overline{AC} = \text{7,3}$ is labelled and $\hat{A} = \text{62,7}°$ . Point $D$ is on side $\overline{AB}$ directly across from point $C$ , such that $\overline{CD}$ makes a right angle with $\overline{AB}$ .

Which of the following statements must be true about the figure above?

Answer: The true statement is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that isosceles triangles are symmetric.

STEP: Consider the choices and select the correct option
[−1 point ⇒ 0 / 1 points left]

The question asks us to select the true statement from the choices in the list. Note that two of the options must be wrong straightaway: $\Delta ABC$ is neither a right-angled triangle nor equilateral. From the remaining options, the correct choice is based on the fact that the segment $\overline{CD}$ divides $\Delta ABC$ into two identical shapes: the segment $\overline{CD}$ is a line of symmetry because the triangle is isosceles. Therefore, point $D$ divides $\overline{AB}$ into two equal parts.

The correct choice from the list is: $\overline{AD}$ = $\overline{DB}$ .

Submit your answer as:
Compute the length of $\overline{AB}$ . Round your answer to one decimal place.

Answer: The length of $\overline{AB}$ is .
numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the segment $\overline{CD}$ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.

STEP: Draw a line to create right-angled triangles
[−1 point ⇒ 2 / 3 points left]

The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

The line segment $\overline{DC}$ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:

STEP: Use trigonometry to find useful information
[−1 point ⇒ 1 / 3 points left]

In $\Delta ADC$ (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

With the information given, we can find both segments $\overline{AD}$ and $\overline{CD}$ . However, for this question only $\overline{AD}$ is useful: we want the length of $\overline{AB}$ , which is twice as long as $\overline{AD}$ . In $\Delta ADC$ , the hypotenuse is $\overline{AC} = \text{7,3}$ , and the side we want is adjacent to the angle given. Therefore we need to use the cosine ratio. Set up the equation and then solve for the length of $\overline{AD}$ .

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (62,7 °) & = \frac{\bar{A D}}{7,3} \\ (7,3) \cos 62,7 ° & = \bar{A D} \\ (7,3) (0,4586...) & = \bar{A D} \\ 3,3481... & = \bar{A D} \end{aligned}$ $\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \left( \text{62,7}° \right) & = \frac{\overline{AD}}{\text{7,3}}\\ \left( \text{7,3} \right) \cos \text{62,7}° & = \overline{AD}\\ \left( \text{7,3} \right) \left(\text{0,4586...}\right) & = \overline{AD}\\ \text{3,3481...} & = \overline{AD} \end{aligned}$

STEP: Calculate the final answer
[−1 point ⇒ 0 / 3 points left]

We want the length of $\overline{AB}$ , so multiply by two since $D$ is the mid-point of segment $\overline{AB}$ :

$\begin{aligned} \bar{A B} & = 2 \cdot \bar{A D} \\ = 2 (3,3481...) \\ = 6,6962... \\ \approx 6,7 \end{aligned}$ $\begin{aligned} \overline{AB} & = 2 \cdot \overline{AD}\\ & = 2 \left( \text{3,3481...} \right)\\ & = \text{6,6962...}\\ & \approx \text{6,7} \end{aligned}$

Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above.

The final answer is: $\overline{AB} = \text{6,7}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point A, looking up at a bird sitting on the top of a building, Point B.

The height of the building is $\text{6}\ \text{m}$ , Point A is $\text{x}\ \text{}$ away from the foot of the building, and the line-of-sight distance from Point A to the top of the building is $\text{7,32}\ \text{m}$ .

Calculate the distance away from the foot of the building ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer: The distance away from the foot of the building is

m

$\text{m}$ .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the Theorem of Pythagoras to calculate the distance away from the foot of the building.

STEP: Use the Theorem of Pythagoras to calculate the distance away from the foot of the building
[−2 points ⇒ 0 / 2 points left]

The question gives us two sides of a right-angled triangle, and we need to find the third side: we can use the Theorem of Pythagoras.

\begin{aligned} c^{2} & = a^{2} + b^{2} \\ {(7, 32)}^{2} & = {(6)}^{2} + x^{2} \\ Solving: x & = 4,19313... \\ \approx 4,2 m \end{aligned}

$\begin{aligned} c^2 & = a^2 + b^2\\ \left( 7,32 \right)^2 & = \left( 6 \right)^2 + x^2\\ \text{Solving: }x & = \text{4,19313...} \\ & \approx \text{4,2}\ \text{m} \end{aligned}$

Therefore, the distance away from the foot of the building is $\text{4,2}\ \text{m}$ .

Submit your answer as:

Exercises

Solving two dimensional problems

A person stands at Point A, looking up at a bird sitting on the top of a building, Point B.

The height of the building is $x$ meters, Point A is $7.0$ meters away from the foot of the building, and the angle of incline (from the horizontal) the the top of the building is $55°$ .

Calculate the height of the building ( $x$ ) as shown in the diagram below. Round your answer to two decimal places where necessary.

Answer:

x

$x$ =

m

$\text{m}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the tangent trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the correct trig ratio to use is the tangent function.

\begin{aligned} \tan (θ) & = \frac{opposite}{adjacent} \\ \tan (55) ° & = \frac{x}{7,0} \\ x & = 7,0 \cdot \tan (55) ° \\ x & = 9,9970... \\ x & \approx 10 \end{aligned}

$\begin{aligned} \tan (\theta) &= \frac{\text{opposite}}{\text{adjacent}}\\ \tan (55)° &= \frac{x}{\text{7,0}}\\ x &= \text{7,0} \cdot \tan(55)° \\ x &= \text{9,9970...}\\ x &\approx 10 \end{aligned}$

Remember to round your answer to two decimal places. Therefore $x$ is $\text{10}\ \text{m}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point A, looking up at a bird sitting on the top of a building, Point B.

The height of the building is $x$ meters, Point A is $5.6$ meters away from the foot of the building, and the angle of incline (from the horizontal) the the top of the building is $55°$ .

Calculate the height of the building ( $x$ ) as shown in the diagram below. Round your answer to two decimal places where necessary.

Answer:

x

$x$ =

m

$\text{m}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the tangent trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the correct trig ratio to use is the tangent function.

\begin{aligned} \tan (θ) & = \frac{opposite}{adjacent} \\ \tan (55) ° & = \frac{x}{5,6} \\ x & = 5,6 \cdot \tan (55) ° \\ x & = 7,9976... \\ x & \approx 8 \end{aligned}

$\begin{aligned} \tan (\theta) &= \frac{\text{opposite}}{\text{adjacent}}\\ \tan (55)° &= \frac{x}{\text{5,6}}\\ x &= \text{5,6} \cdot \tan(55)° \\ x &= \text{7,9976...}\\ x &\approx 8 \end{aligned}$

Remember to round your answer to two decimal places. Therefore $x$ is $\text{8}\ \text{m}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point A, looking up at a bird sitting on the top of a building, Point B.

The height of the building is $x$ meters, Point A is $4.04$ meters away from the foot of the building, and the angle of incline (from the horizontal) the the top of the building is $60°$ .

Calculate the height of the building ( $x$ ) as shown in the diagram below. Round your answer to two decimal places where necessary.

Answer:

x

$x$ =

m

$\text{m}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the tangent trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the correct trig ratio to use is the tangent function.

\begin{aligned} \tan (θ) & = \frac{opposite}{adjacent} \\ \tan (60) ° & = \frac{x}{4,04} \\ x & = 4,04 \cdot \tan (60) ° \\ x & = 6,9974... \\ x & \approx 7 \end{aligned}

$\begin{aligned} \tan (\theta) &= \frac{\text{opposite}}{\text{adjacent}}\\ \tan (60)° &= \frac{x}{\text{4,04}}\\ x &= \text{4,04} \cdot \tan(60)° \\ x &= \text{6,9974...}\\ x &\approx 7 \end{aligned}$

Remember to round your answer to two decimal places. Therefore $x$ is $\text{7}\ \text{m}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point $A$ , looking up at a bird sitting on the top of a building, at Point $B$ .

The height of the building is $\text{9}\ \text{m}$ , the line-of-sight distance from Point $A$ to the top of the building (Point $B$ ) is $x$ , and the angle of inclination (from the horizontal) to the top of the building is $\text{52}°$ .

Calculate the line-of-sight distance to the top of the building ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer: The line-of-sight distance to the top of the building is

m

$\text{m}$ .

one-of

type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the sine trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the trig ratio to use is $\sin$ .

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin 52 ° & = \frac{9}{x} \\ x & = \frac{9}{\sin 52 °} \\ x & = 11,42116... \\ \approx 11,42 m \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin \text{52}\text{°} & = \frac{\text{9}}{x} \\ x & = \frac{\text{9}}{\sin\text{52}\text{°}}\\ x & = \text{11,42116...}\\ & \approx \text{11,42}\ \text{m} \end{aligned}$

The line-of-sight distance to the top of the building is $\text{11,42}\ \text{m}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point $A$ , looking up at a bird sitting on the top of a building, at Point $B$ .

The height of the building is $\text{5}\ \text{m}$ , the line-of-sight distance from Point $A$ to the top of the building (Point $B$ ) is $\text{5,72}\ \text{m}$ , and the angle of inclination (from the horizontal) to the top of the building is $x$ .

Calculate the angle of inclination (from the horizontal) to the top of the building ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer: The angle of inclination (from the horizontal) to the top of the building is

°

$\text{°}$ .

one-of

type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the sine trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the trig ratio to use is $\sin$ .

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin x & = \frac{5}{5,72} \\ x & = \sin^{- 1} (\frac{5}{5,72}) \\ x & = 60,94169... \\ \approx 60,94 m \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin x & = \frac{\text{5}}{\text{5,72}} \\ x & = \sin^{-1} \left(\frac{\text{5}}{\text{5,72}}\right)\\ x & = \text{60,94169...}\\ & \approx \text{60,94}\ \text{m} \end{aligned}$

The angle of inclination (from the horizontal) to the top of the building is $\text{60,94}\ \text{m}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point $A$ , looking up at a bird sitting on the top of a tree, at Point $B$ .

The height of the tree is $\text{10}\ \text{m}$ , the line-of-sight distance from Point $A$ to the top of the tree (Point $B$ ) is $\text{11,03}\ \text{m}$ , and the angle of inclination (from the horizontal) to the top of the tree is $x$ .

Calculate the angle of inclination (from the horizontal) to the top of the tree ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer: The angle of inclination (from the horizontal) to the top of the tree is

°

$\text{°}$ .

one-of

type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the sine trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the trig ratio to use is $\sin$ .

\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin x & = \frac{10}{11,03} \\ x & = \sin^{- 1} (\frac{10}{11,03}) \\ x & = 65,04213... \\ \approx 65,04 m \end{aligned}

$\begin{aligned} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin x & = \frac{\text{10}}{\text{11,03}} \\ x & = \sin^{-1} \left(\frac{\text{10}}{\text{11,03}}\right)\\ x & = \text{65,04213...}\\ & \approx \text{65,04}\ \text{m} \end{aligned}$

The angle of inclination (from the horizontal) to the top of the tree is $\text{65,04}\ \text{m}$ .

Submit your answer as:

Using trigonometry with word problems

At a house in Polokwane, a father stands at the top of the stairs. His child is at the bottom of the stairs. The following are true:

At the bottom of the stairs, there is an angle of $\text{46,4}\text{°}$ between the slope of the stairs and the vertical direction.
The horizontal distance from the father at the top of the stairs to the point straight up from the child is $\text{4,2}\ \text{m}$ .

Calculate the vertical distance of the father above the child (the height from the top of the stairs to the bottom).

INSTRUCTION: Round your answer to two decimal places.

Answer: The distance is metres.

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

STEP: Draw a diagram to represent the situation and label it
[−1 point ⇒ 3 / 4 points left]

When you have a word problem about a shape or positions and there is no diagram, start by drawing a sketch to summarise the question. Label the sketch with all of the information given. In this case, the information given in the question fits on a triangle: there is one thing at ground level (the child at the bottom of the stairs) and something else up above (the father at the top of the stairs). We can represent this with a right-angled triangle.

The picture does not need to be perfect but try to make it reasonable. For example, this question gives an angle of $\text{46,4}\text{°}$ : try to make the angle in the sketch reasonably close to that size.

NOTE: The diagram should also have a label for what you want to find. We have used "

y

$y$ " to label the distance that we want to find (but you can use any variable that you want).

STEP: Determine which trigonometric ratio to use
[−1 point ⇒ 2 / 4 points left]

One of the most important facts about the picture above is that it is a right-angled triangle: we can draw a vertical line straight up from the child, which is perpendicular to the horizontal line shown. That means we can use the trigonometric ratios to try to solve this question, or even the theorem of Pythagoras: we should pick whichever option will get us the answer.

In this case, we do not know anything about the hypotenuse. Instead, with respect to the angle given, we know the opposite side and we need to determine the length of the adjacent side. Therefore, the correct trigonometric ratio is the tangent function. (In fact, there is no other way to solve the question with the information given. While the theorem of Pythagoras applies to the triangle, we do not know enough information to use it for this question.)

STEP: Set up an equation and solve for the answer
[−2 points ⇒ 0 / 4 points left]

Now we can write an equation for the tangent ratio from the sketch, and solve for the distance we need. Start by writing the equation for tangent, and then substitute in the correct values.

\begin{aligned} \tan θ & = \frac{opposite}{adjacent} \\ \tan (46,4 °) & = \frac{4,2}{y} \\ y \times \tan 46,4 ° & = \frac{4,2}{y} \times y \leftarrow_{to clear the denominator}^{Multiply both sides by y} \\ y & = \frac{4,2}{\tan 46,4 °} \\ y & = 3,9996... \\ y & \approx 4 m \end{aligned}

$\begin{aligned} \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} \\ \tan \left( \text{46,4}° \right) & = \frac{\text{4,2}}{y}\\ y \times \tan \text{46,4}° & = \frac{\text{4,2}}{y} \times y \quad \leftarrow \quad ^{\text{Multiply both sides by y}}_{\text{to clear the denominator}}\\ y & = \frac{\text{4,2}}{\tan \text{46,4}°} \\ y & = \text{3,9996...} \\ y & \approx \text{4}\ \text{m} \end{aligned}$

We got it! As shown in the last step, we rounded off to the nearest hundredths place because the question told us to. (In this case, rounding off to two decimal places results in an integer value. If you give the answer as $\text{4,00}$ that is fine.)

The vertical distance of the father above the child (the height from the top of the stairs to the bottom) is $\text{4}\ \text{metres}$ .

Submit your answer as:

Using trigonometry with word problems

A man stands at a window on the first floor of a building, shouting down to a child on the other side of the street. Here are two facts about the situation:

For the child on the street, the line to the man in the window makes an angle of $\text{60}\text{°}$ with the vertical direction.
The man in the window is $\text{3}\ \text{m}$ above the street.

Find the oblique (slanted) distance between the man and the child.

INSTRUCTION: Round your answer to two decimal places.

Answer: The distance is metres.

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

STEP: Draw a diagram to represent the situation and label it
[−1 point ⇒ 3 / 4 points left]

The picture does not need to be perfect but try to make it reasonable. For example, this question gives an angle of $\text{60}\text{°}$ : try to make the angle in the sketch reasonably close to that size.

NOTE: The diagram should also have a label for what you want to find. We have used "

h

$h$ " to label the distance that we want to find (but you can use any variable that you want).

STEP: Determine which trigonometric ratio to use
[−1 point ⇒ 2 / 4 points left]

In this question we need to find the hypotenuse of the triangle, and the side that we already know is adjacent to the angle given. Therefore, the correct trigonometric ratio is the cosine function. (In fact, there is no other way to solve the question with the information given. While the theorem of Pythagoras applies to the triangle, we do not know enough information to use it for this question.)

STEP: Set up an equation and solve for the answer
[−2 points ⇒ 0 / 4 points left]

Now we can write an equation for the cosine ratio from the sketch, and solve for the distance we need. Start by writing the equation for cosine, and then substitute in the correct values.

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (60 °) & = \frac{3}{h} \\ h \times \cos 60 ° & = \frac{3}{h} \times h \leftarrow_{to clear the denominator}^{Multiply both sides by h} \\ h & = \frac{3}{\cos 60 °} \\ h & = 6 \\ h & \approx 6 m \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \\ \cos \left( \text{60}° \right) & = \frac{\text{3}}{h}\\ h \times \cos \text{60}° & = \frac{\text{3}}{h} \times h \quad \leftarrow \quad ^{\text{Multiply both sides by h}}_{\text{to clear the denominator}}\\ h & = \frac{\text{3}}{\cos \text{60}°} \\ h & = \text{6} \\ h & \approx \text{6}\ \text{m} \end{aligned}$

The oblique (slanted) distance between the man and the child is $\text{6}\ \text{metres}$ .

Submit your answer as:

Using trigonometry with word problems

Imagine a cat sitting up in a tree watching a mouse on the ground. Consider these two statements about the situation:

The distance of the cat above the ground is $\text{7}\ \text{m}$ .
The angle at the cat between the mouse and the vertical direction is $\text{16,6}\text{°}$ .

Determine the oblique (slanted) distance between the mouse and the cat.

INSTRUCTION: Round your answer to two decimal places.

Answer: The distance is metres.

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

STEP: Draw a diagram to represent the situation and label it
[−1 point ⇒ 3 / 4 points left]

When you have a word problem about a shape or positions and there is no diagram, start by drawing a sketch to summarise the question. Label the sketch with all of the information given. In this case, the information given in the question fits on a triangle: there is one thing at ground level (the mouse on the ground) and something else up above (the cat in the tree). We can represent this with a right-angled triangle.

The picture does not need to be perfect but try to make it reasonable. For example, this question gives an angle of $\text{16,6}\text{°}$ : try to make the angle in the sketch reasonably close to that size.

NOTE: The diagram should also have a label for what you want to find. We have used "

h

$h$ " to label the distance that we want to find (but you can use any variable that you want).

STEP: Determine which trigonometric ratio to use
[−1 point ⇒ 2 / 4 points left]

One of the most important facts about the picture above is that it is a right-angled triangle: we can draw a vertical line straight down from the cat, which is perpendicular to the ground. That means we can use the trigonometric ratios to try to solve this question, or even the theorem of Pythagoras: we should pick whichever option will get us the answer.

STEP: Set up an equation and solve for the answer
[−2 points ⇒ 0 / 4 points left]

Now we can write an equation for the cosine ratio from the sketch, and solve for the distance we need. Start by writing the equation for cosine, and then substitute in the correct values.

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (16,6 °) & = \frac{7}{h} \\ h \times \cos 16,6 ° & = \frac{7}{h} \times h \leftarrow_{to clear the denominator}^{Multiply both sides by h} \\ h & = \frac{7}{\cos 16,6 °} \\ h & = 7,3044... \\ h & \approx 7,3 m \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \\ \cos \left( \text{16,6}° \right) & = \frac{\text{7}}{h}\\ h \times \cos \text{16,6}° & = \frac{\text{7}}{h} \times h \quad \leftarrow \quad ^{\text{Multiply both sides by h}}_{\text{to clear the denominator}}\\ h & = \frac{\text{7}}{\cos \text{16,6}°} \\ h & = \text{7,3044...} \\ h & \approx \text{7,3}\ \text{m} \end{aligned}$

We got it! As shown in the last step, we rounded off to the nearest hundredths place because the question told us to. (In this case, rounding off to two decimal places leads to a number which stops after the first decimal place. However, you can give your answer as $\text{7,30}$ if you want to.)

The oblique (slanted) distance between the mouse and the cat is $\text{7,3}\ \text{metres}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point $A$ , looking up at a bird sitting on the top of a tree, Point $B$ .

Point $A$ is $7.0$ meters away from the foot of the tree, the line-of-sight distance from Point $A$ to the top of the tree (Point $B$ ) is $\text{9,9}$ meters, and the angle of incline (from the horizontal) the the top of the tree is $\text{x}\text{°}$ .

Calculate the angle of incline (from the horizontal) the the top of the tree ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x=$

°

$\text{°}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the cosine trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the trig ratio to use is $\cos$ .

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos x ° & = \frac{7.0}{9,9} \\ ∴ x & = 45 ° \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \text{x}° & = \frac{7.0}{\text{9,9}} \\ \therefore x & = \text{45}\text{°} \end{aligned}$

Therefore $x$ is $\text{45}\text{°}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point $A$ , looking up at a bird sitting on the top of a tree, Point $B$ .

Calculate the angle of incline (from the horizontal) the the top of the tree ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x=$

°

$\text{°}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the cosine trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the trig ratio to use is $\cos$ .

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos x ° & = \frac{7.0}{9,9} \\ ∴ x & = 45 ° \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \text{x}° & = \frac{7.0}{\text{9,9}} \\ \therefore x & = \text{45}\text{°} \end{aligned}$

Therefore $x$ is $\text{45}\text{°}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point $A$ , looking up at a bird sitting on the top of a building, Point $B$ .

Point $A$ is $4.2$ meters away from the foot of the building, the line-of-sight distance from Point $A$ to the top of the building (Point $B$ ) is $x$ meters, and the angle of incline (from the horizontal) the the top of the building is $\text{55}\text{°}$ .

Calculate the line-of-sight distance to the top of the building ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer:

x =

$x=$

m

$\text{m}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Identify the opposite, adjacent and hypotenuse sides and use the appropriate trigonometric ratio to solve for $x$ .

STEP: Use the cosine trigonometric ratio to solve for

x

$x$
[−2 points ⇒ 0 / 2 points left]

First identify the opposite, adjacent and hypotenuse sides. Then use the appropriate trigonometric ratio to solve for $x$ . In this example, the trig ratio to use is $\cos$ .

\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos 55 ° & = \frac{4.2}{x} \\ ∴ x & = 7,32 m \end{aligned}

$\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \text{55}° & = \frac{4.2}{x} \\ \therefore x & = \text{7,32}\ \text{m} \end{aligned}$

Therefore $x$ is $\text{7,32}\ \text{m}$ .

Submit your answer as:

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows a scalene triangle. Two sides and an angle are given: $\overline{DF} = \text{8,9}$ , $\; \overline{DE} = \text{9}$ and $\hat{D} = \text{52,4}°$ . Point $G$ is on side $\overline{DE}$ as labelled, such that $\overline{FG}$ makes a right angle with $\overline{DE}$ .

Which of the following statements must be true about the figure above?

Answer: The true statement is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Imagine walking from point $D$ to $G$ and then walking on from $G$ to $E$ . How does this compare to walking straight from $D$ to $E$ without stopping?

STEP: Consider the choices and select the correct option
[−1 point ⇒ 0 / 1 points left]

The question asks us to select the true statement from the choices in the list. The correct option is about the sum of the segments along side $\overline{DE}$ . Point $G$ breaks $\overline{DE}$ up into two pieces. Therefore we know that together they make the total length of $\overline{DE}$ . Note that we do not know the exact position of point $G$ , only that it is somewhere between $D$ and $E$ .

The correct choice from the list is: Side $DE$ is the sum of sides $DG$ and $GE$ .

Submit your answer as:
Now determine the length of $\overline{EF}$ . Round your answer to one decimal place.

Answer: The length of $\overline{EF}$ is .
numeric

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Use the segment $\overline{FG}$ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.

STEP: Draw a line to create right-angled triangles
[−1 point ⇒ 6 / 7 points left]

The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

The line segment $\overline{GF}$ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:

STEP: Use trigonometry to find useful information
[−4 points ⇒ 2 / 7 points left]

In $\Delta DGF$ (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

With the information given, we can find both segments $\overline{DG}$ and $\overline{FG}$ . Remember that we want to get the length of $\overline{EF}$ , and for that we need both of these lengths. Start by calculating the length of $\overline{DG}$ , which will allow us to find $\overline{GE}$ (because we know that $\overline{DE} = \text{9}$ ). This calculation involves the hypotenuse and the side adjacent to $\hat{D}$ , so use the cosine ratio.

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (52,4 °) & = \frac{\bar{D G}}{8,9} \\ (8,9) \cos 52,4 ° & = \bar{D G} \\ (8,9) (0,6101...) & = \bar{D G} \\ 5,4302... & = \bar{D G} \\ Since \bar{D G} + \bar{G E} = \bar{D E} : \\ \bar{G E} = 9 - 5,4302... \\ \bar{G E} = 3,5697... \end{aligned}$ $\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \left( \text{52,4}° \right) & = \frac{\overline{DG}}{\text{8,9}}\\ \left( \text{8,9} \right) \cos \text{52,4}° & = \overline{DG}\\ \left( \text{8,9} \right) \left(\text{0,6101...}\right) & = \overline{DG}\\ \text{5,4302...} & = \overline{DG}\\ \text{Since} \quad \overline{DG} + \overline{GE} = \overline{DE}: &\\ & \overline{GE} = \text{9} - \text{5,4302...}\\ & \overline{GE} = \text{3,5697...} \end{aligned}$

Great: that gets us the value for side $\overline{GE}$ . Now we need to find the length of side $\overline{FG}$ . For that we will use the sine ratio (you can also do this calculation with the theorem of Pythagoras, but here we will do it with trigonometry).

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin (52,4 °) & = \frac{\bar{F G}}{8,9} \\ (8,9) \sin 52,4 ° & = \bar{F G} \\ (8,9) (0,7922...) & = \bar{F G} \\ 7,0513... & = \bar{F G} \end{aligned}$ $\begin{aligned}\sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin \left( \text{52,4}° \right) & = \frac{\overline{FG}}{\text{8,9}}\\ \left( \text{8,9} \right) \sin \text{52,4}° & = \overline{FG}\\ \left( \text{8,9} \right) \left(\text{0,7922...}\right) & = \overline{FG}\\ \text{7,0513...} & = \overline{FG} \end{aligned}$

STEP: Calculate the final answer
[−2 points ⇒ 0 / 7 points left]

Now we have the lengths of two sides of $\Delta GEF$ ; since $\Delta GEF$ is a right-angled triangle, we can use the theorem of Pythagoras to calculate the length of the hypotenuse.

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ {(\bar{E F})}^{2} & = {(\bar{F G})}^{2} + {(\bar{G E})}^{2} \\ = {(7,0513...)}^{2} + {(3,5697...)}^{2} \\ = 62,4647... \\ ∴ \bar{E F} & = \pm \sqrt{62,4647...} \leftarrow_{the \pm because a^{2} = (- a)^{2}}^{The square root brings in} \\ = \pm 7,9034... \\ \approx \pm 7,9 \end{aligned}$ $\begin{aligned} c^{2} & = a^{2} + b^{2}\\ \left( \overline{EF} \right)^{2} & = \left( \overline{FG} \right)^{2} + \left( \overline{GE} \right)^{2}\\ & =\left( \text{7,0513...} \right)^{2} + \left( \text{3,5697...} \right)^{2}\\ & = \text{62,4647...}\\ \therefore \overline{EF} & = \pm \sqrt{\text{62,4647...}} \quad \leftarrow \quad ^{\text{The square root brings in}}_{\text{the } \pm \text{ because } a^{2} = (-a)^{2}}\\ & = \pm \text{7,9034...}\\ & \approx \pm \text{7,9} \end{aligned}$

Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above. Also notice that we get two different answers, one positive and one negative. However, the value we calculated represents a distance, so we must throw out the negative answer.

The final answer is: $\overline{EF} = \text{7,9}$ .

Submit your answer as:

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows a scalene triangle. Two sides and an angle are given: $\overline{PR} = \text{5}$ , $\; \overline{PQ} = \text{6,1}$ and $\hat{P} = \text{38,5}°$ . Point $S$ is on side $\overline{PQ}$ as labelled, such that $\overline{RS}$ makes a right angle with $\overline{PQ}$ .

Which of the following statements must be true about the figure above?

Answer: The true statement is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Imagine walking from point $P$ to $S$ and then walking on from $S$ to $Q$ . How does this compare to walking straight from $P$ to $Q$ without stopping?

STEP: Consider the choices and select the correct option
[−1 point ⇒ 0 / 1 points left]

The question asks us to select the true statement from the choices in the list. The correct option is about the sum of the segments along side $\overline{PQ}$ . Point $S$ breaks $\overline{PQ}$ up into two pieces. Therefore we know that together they make the total length of $\overline{PQ}$ . Note that we do not know the exact position of point $S$ , only that it is somewhere between $P$ and $Q$ .

The correct choice from the list is: Side $PQ$ is the sum of sides $PS$ and $SQ$ .

Submit your answer as:
Now determine the length of $\overline{QR}$ . Round your answer to one decimal place.

Answer: The length of $\overline{QR}$ is .
numeric

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Use the segment $\overline{RS}$ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.

STEP: Draw a line to create right-angled triangles
[−1 point ⇒ 6 / 7 points left]

The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

The line segment $\overline{SR}$ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:

STEP: Use trigonometry to find useful information
[−4 points ⇒ 2 / 7 points left]

In $\Delta PSR$ (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

With the information given, we can find both segments $\overline{PS}$ and $\overline{RS}$ . Remember that we want to get the length of $\overline{QR}$ , and for that we need both of these lengths. Start by calculating the length of $\overline{PS}$ , which will allow us to find $\overline{SQ}$ (because we know that $\overline{PQ} = \text{6,1}$ ). This calculation involves the hypotenuse and the side adjacent to $\hat{P}$ , so use the cosine ratio.

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (38,5 °) & = \frac{\bar{P S}}{5} \\ (5) \cos 38,5 ° & = \bar{P S} \\ (5) (0,7826...) & = \bar{P S} \\ 3,9130... & = \bar{P S} \\ Since \bar{P S} + \bar{S Q} = \bar{P Q} : \\ \bar{S Q} = 6,1 - 3,9130... \\ \bar{S Q} = 2,1869... \end{aligned}$ $\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \left( \text{38,5}° \right) & = \frac{\overline{PS}}{\text{5}}\\ \left( \text{5} \right) \cos \text{38,5}° & = \overline{PS}\\ \left( \text{5} \right) \left(\text{0,7826...}\right) & = \overline{PS}\\ \text{3,9130...} & = \overline{PS}\\ \text{Since} \quad \overline{PS} + \overline{SQ} = \overline{PQ}: &\\ & \overline{SQ} = \text{6,1} - \text{3,9130...}\\ & \overline{SQ} = \text{2,1869...} \end{aligned}$

Great: that gets us the value for side $\overline{SQ}$ . Now we need to find the length of side $\overline{RS}$ . For that we will use the sine ratio (you can also do this calculation with the theorem of Pythagoras, but here we will do it with trigonometry).

$\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} \\ \sin (38,5 °) & = \frac{\bar{R S}}{5} \\ (5) \sin 38,5 ° & = \bar{R S} \\ (5) (0,6225...) & = \bar{R S} \\ 3,1125... & = \bar{R S} \end{aligned}$ $\begin{aligned}\sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}}\\ \sin \left( \text{38,5}° \right) & = \frac{\overline{RS}}{\text{5}}\\ \left( \text{5} \right) \sin \text{38,5}° & = \overline{RS}\\ \left( \text{5} \right) \left(\text{0,6225...}\right) & = \overline{RS}\\ \text{3,1125...} & = \overline{RS} \end{aligned}$

STEP: Calculate the final answer
[−2 points ⇒ 0 / 7 points left]

Now we have the lengths of two sides of $\Delta SQR$ ; since $\Delta SQR$ is a right-angled triangle, we can use the theorem of Pythagoras to calculate the length of the hypotenuse.

$\begin{aligned} c^{2} & = a^{2} + b^{2} \\ {(\bar{Q R})}^{2} & = {(\bar{R S})}^{2} + {(\bar{S Q})}^{2} \\ = {(3,1125...)}^{2} + {(2,1869...)}^{2} \\ = 14,4709... \\ ∴ \bar{Q R} & = \pm \sqrt{14,4709...} \leftarrow_{the \pm because a^{2} = (- a)^{2}}^{The square root brings in} \\ = \pm 3,8040... \\ \approx \pm 3,8 \end{aligned}$ $\begin{aligned} c^{2} & = a^{2} + b^{2}\\ \left( \overline{QR} \right)^{2} & = \left( \overline{RS} \right)^{2} + \left( \overline{SQ} \right)^{2}\\ & =\left( \text{3,1125...} \right)^{2} + \left( \text{2,1869...} \right)^{2}\\ & = \text{14,4709...}\\ \therefore \overline{QR} & = \pm \sqrt{\text{14,4709...}} \quad \leftarrow \quad ^{\text{The square root brings in}}_{\text{the } \pm \text{ because } a^{2} = (-a)^{2}}\\ & = \pm \text{3,8040...}\\ & \approx \pm \text{3,8} \end{aligned}$

Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above. Also notice that we get two different answers, one positive and one negative. However, the value we calculated represents a distance, so we must throw out the negative answer.

The final answer is: $\overline{QR} = \text{3,8}$ .

Submit your answer as:

Trigonometry with non-right triangles

The figure below, which is drawn to scale, shows an isosceles triangle. Side $\overline{VX} = \text{7,1}$ is labelled and $\hat{V} = \text{58,6}°$ . Point $Y$ is on side $\overline{VW}$ directly across from point $X$ , such that $\overline{XY}$ makes a right angle with $\overline{VW}$ .

Which of the following statements must be true about the figure above?

Answer: The true statement is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that isosceles triangles are symmetric.

STEP: Consider the choices and select the correct option
[−1 point ⇒ 0 / 1 points left]

The question asks us to select the true statement from the choices in the list. Note that two of the options must be wrong straightaway: $\Delta VWX$ is neither a right-angled triangle nor equilateral. From the remaining options, the correct choice is based on the fact that the segment $\overline{XY}$ divides $\Delta VWX$ into two identical shapes: the segment $\overline{XY}$ is a line of symmetry because the triangle is isosceles. Therefore, point $Y$ divides $\overline{VW}$ into two equal parts.

The correct choice from the list is: Point $Y$ divides side $VW$ into two equal pieces.

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Next, find the length of $\overline{VW}$ . Round your answer to one decimal place.

Answer: The length of $\overline{VW}$ is .
numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the segment $\overline{XY}$ to get two right-angled triangles. Then you can use trigonometric ratios or the theorem of Pythagoras to work out the answer to the question.

STEP: Draw a line to create right-angled triangles
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The first thing to do is draw an extra line across the triangle so that we make two right-angled triangles in the figure. We do this because we can use the trigonometric ratios and the theorem of Pythagoras for right-angled triangles.

The line segment $\overline{YX}$ in the figure is the line we want: it will create two separate right-angled triangles! The two right-angled triangles that we get look like this:

STEP: Use trigonometry to find useful information
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In $\Delta VYX$ (the light blue one) we know one of the non-right angles and one of the sides. Hence we can use the trigonometric ratios in that triangle because it is a right-angled triangle.

With the information given, we can find both segments $\overline{VY}$ and $\overline{XY}$ . However, for this question only $\overline{VY}$ is useful: we want the length of $\overline{VW}$ , which is twice as long as $\overline{VY}$ . In $\Delta VYX$ , the hypotenuse is $\overline{VX} = \text{7,1}$ , and the side we want is adjacent to the angle given. Therefore we need to use the cosine ratio. Set up the equation and then solve for the length of $\overline{VY}$ .

$\begin{aligned} \cos θ & = \frac{adjacent}{hypotenuse} \\ \cos (58,6 °) & = \frac{\bar{V Y}}{7,1} \\ (7,1) \cos 58,6 ° & = \bar{V Y} \\ (7,1) (0,5210...) & = \bar{V Y} \\ 3,6991... & = \bar{V Y} \end{aligned}$ $\begin{aligned} \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \left( \text{58,6}° \right) & = \frac{\overline{VY}}{\text{7,1}}\\ \left( \text{7,1} \right) \cos \text{58,6}° & = \overline{VY}\\ \left( \text{7,1} \right) \left(\text{0,5210...}\right) & = \overline{VY}\\ \text{3,6991...} & = \overline{VY} \end{aligned}$

STEP: Calculate the final answer
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We want the length of $\overline{VW}$ , so multiply by two since $Y$ is the mid-point of segment $\overline{VW}$ :

$\begin{aligned} \bar{V W} & = 2 \cdot \bar{V Y} \\ = 2 (3,6991...) \\ = 7,3983... \\ \approx 7,4 \end{aligned}$ $\begin{aligned} \overline{VW} & = 2 \cdot \overline{VY}\\ & = 2 \left( \text{3,6991...} \right)\\ & = \text{7,3983...}\\ & \approx \text{7,4} \end{aligned}$

Remember that the instructions say to round the answer to the first decimal place, as shown in the last step above.

The final answer is: $\overline{VW} = \text{7,4}$ .

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Solving two dimensional problems

A person stands at Point A, looking up at a bird sitting on the top of a building, Point B.

The height of the building is $\text{7}\ \text{m}$ , Point A is $\text{8,34}\ \text{m}$ away from the foot of the building, and the line-of-sight distance from Point A to the top of the building is $\text{x}\ \text{}$ .

Calculate the line-of-sight distance to the top of the building ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer: The line-of-sight distance to the top of the building is

m

$\text{m}$ .

numeric

HINT: <no title>
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Use the Theorem of Pythagoras to calculate the line-of-sight distance to the top of the building.

STEP: Use the Theorem of Pythagoras to calculate the line-of-sight distance to the top of the building
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The question gives us two sides of a right-angled triangle, and we need to find the third side: we can use the Theorem of Pythagoras.

\begin{aligned} c^{2} & = a^{2} + b^{2} \\ x^{2} & = {(7)}^{2} + {(8, 34)}^{2} \\ Solving: x & = 10,88832... \\ \approx 10,89 m \end{aligned}

$\begin{aligned} c^2 & = a^2 + b^2\\ x^2 & = \left( 7 \right)^2 + \left( 8,34 \right)^2\\ \text{Solving: }x & = \text{10,88832...} \\ & \approx \text{10,89}\ \text{m} \end{aligned}$

Therefore, the line-of-sight distance to the top of the building is $\text{10,89}\ \text{m}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point A, looking up at a bird sitting on the top of a building, Point B.

The height of the building is $\text{x}\ \text{}$ , Point A is $\text{5,2}\ \text{m}$ away from the foot of the building, and the line-of-sight distance from Point A to the top of the building is $\text{10,39}\ \text{m}$ .

Calculate the height of the building ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer: The height of the building is

m

$\text{m}$ .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the Theorem of Pythagoras to calculate the height of the building.

STEP: Use the Theorem of Pythagoras to calculate the height of the building
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The question gives us two sides of a right-angled triangle, and we need to find the third side: we can use the Theorem of Pythagoras.

\begin{aligned} c^{2} & = a^{2} + b^{2} \\ {(10, 39)}^{2} & = x^{2} + {(5, 2)}^{2} \\ Solving: x & = 8,99511... \\ \approx 9 m \end{aligned}

$\begin{aligned} c^2 & = a^2 + b^2\\ \left( 10,39 \right)^2 & = x^2 + \left( 5,2 \right)^2\\ \text{Solving: }x & = \text{8,99511...} \\ & \approx \text{9}\ \text{m} \end{aligned}$

Therefore, the height of the building is $\text{9}\ \text{m}$ .

Submit your answer as:

Solving two dimensional problems

A person stands at Point A, looking up at a bird sitting on the top of a building, Point B.

The height of the building is $\text{9}\ \text{m}$ , Point A is $\text{x}\ \text{}$ away from the foot of the building, and the line-of-sight distance from Point A to the top of the building is $\text{14,0}\ \text{m}$ .

Calculate the distance away from the foot of the building ( $x$ ) as shown in the diagram below:

INSTRUCTION: Round your answer to two decimal places.

Answer: The distance away from the foot of the building is

m

$\text{m}$ .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the Theorem of Pythagoras to calculate the distance away from the foot of the building.

STEP: Use the Theorem of Pythagoras to calculate the distance away from the foot of the building
[−2 points ⇒ 0 / 2 points left]

The question gives us two sides of a right-angled triangle, and we need to find the third side: we can use the Theorem of Pythagoras.

\begin{aligned} c^{2} & = a^{2} + b^{2} \\ {(14, 0)}^{2} & = {(9)}^{2} + x^{2} \\ Solving: x & = 10,72380... \\ \approx 10,73 m \end{aligned}

$\begin{aligned} c^2 & = a^2 + b^2\\ \left( 14,0 \right)^2 & = \left( 9 \right)^2 + x^2\\ \text{Solving: }x & = \text{10,72380...} \\ & \approx \text{10,73}\ \text{m} \end{aligned}$

Therefore, the distance away from the foot of the building is $\text{10,73}\ \text{m}$ .

Submit your answer as: